A roller-coaster car of mass 1.50 × 10 3 kg is initially at the top of a rise at point . It then moves 35.0 m at an angle of 50.0° below the horizontal to a lower point . (a) Find both the potential energy of the system when the car is at points and and the change in potential energy as the car moves from point to point , assuming y = 0 at point . (b) Repeat part (a), this time choosing y = 0 at point , which is another 15.0 m down the same slope from point .

Question

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Answer 1

Textbook Question

Chapter 5, Problem 63AP

A roller-coaster car of mass 1.50 × 10³ kg is initially at the top of a rise at point Chapter 5, Problem 63AP, A roller-coaster car of mass 1.50 103 kg is initially at the top of a rise at point . It then moves , example 1 . It then moves 35.0 m at an angle of 50.0° below the horizontal to a lower point . (a) Find both the potential energy of the system when the car is at points and Chapter 5, Problem 63AP, A roller-coaster car of mass 1.50 103 kg is initially at the top of a rise at point . It then moves , example 4 and the change in potential energy as the car moves from point to point , assuming y = 0 at point . (b) Repeat part (a), this time choosing y = 0 at point , which is another 15.0 m down the same slope from point Chapter 5, Problem 63AP, A roller-coaster car of mass 1.50 103 kg is initially at the top of a rise at point . It then moves , example 9 .

(a)

Expert Solution

To determine

Bothe the potential energy of the system when the car is on A and B and the change in potential energy of the car when it moves from A to B, when

$y=0$ at point B.

Answer to Problem 63AP

The potential energy of the system at point A is

$3.94×105 J$ .

The potential energy of the system on point B is zero.

The change in potential energy of the car when it moves from A to B is $−3.94×105 J$ when $y=0$ at point B.

Explanation of Solution

Section1:

To determine: The potential energy of the system at point A.

Answer: the potential energy of the system at point A is $3.94×105 J$ .

Explanation:

Given Info:

The mass of the roller-coaster car is $1.50×103 kg$ .

The distance that the car travelled between point A and B is $35.0 m$ .

The path is at an angle $50.0°$ below the horizontal to the lower point B.

Since $y=0$ is considered at point B,

$yB=0$

$yA=lsinθ$ (I)

$yB$ is the height to the point B from the point $y=0$
$yA$ is the height to the point A from the point $y=0$
l is the length that the car moves
$θ$ is the angle of the path of the car to the horizontal

Formula to calculate the potential energy of the system at point A is,

$PEA=mgyA$ (II)

m is mass of the car
g is acceleration due to gravity

Substitute equation (I) in (II),

$PEA=mg(lsinθ)$

Substitute $1.50×103 kg$ for $m$ , $9.8 m/s2$ for g, $35.0 m$ for l and $50.0°$ for $θ$ to find the potential energy of the system at point A,

$PEA=(1.50×103 kg)(9.8 m/s2)((35.0 m)sin50.0°)=3.94×105 J$

Thus, the potential energy of the system at point A is $3.94×105 J$ .

Section 2:

To determine: The potential energy of the system on point B

Answer: The potential energy of the system on point B is zero.

Explanation:

Since $y=0$ is considered at point B, the potential energy of the system at point B is zero.

Conclusion:

From section1,

The potential energy of the system at point A is $3.94×105 J$ .

From section2,

The potential energy of the system on point B is zero.

Thus,

The change in potential energy when the system travels from A to B is,

$ΔPEA→B=PEB−PEA$

Substitute zero for $PEB$ and $3.94×105 J$ for $PEA$ to find the change in kinetic energy,

$ΔPEA→B=0−3.94×105 J=−3.9×105 J$

Therefore, the change in potential energy of the car when it moves from A to B is $−3.9×105 J$ when $y=0$ at point B.

(b)

Expert Solution

To determine

Bothe the potential energy of the system when the car is on A and B and the change in potential energy of the car when it moves from A to B, when

$y=0$ at point C which is

$15.0 m$ down the slope from point B.

Answer to Problem 63AP

The potential energy of the system at point A is

$5.63×105 J$ .

The potential energy of the system on point B is $1.69×105 J$ .

The change in potential energy of the car when it moves from A to B is $−3.94×105 J$ when $y=0$ at point C which is $15.0 m$ down the slope from point B.

Explanation of Solution

Section1:

To determine: The potential energy of the system at point A.

Answer: the potential energy of the system at point A is $5.63×105 J$ .

Explanation:

Given Info:

The mass of the roller-coaster car is $1.50×103 kg$ .

The distance that the car travelled between point A and C is $50.0 m$ .

The path is at an angle $50.0°$ below the horizontal to the lower point B.

Since $y=0$ is considered at point C,

$yB=lsinθ$ (I)

$yA=lsinθ$ (II)

$yB$ is the height to the point B from the point $y=0$
$yA$ is the height to the point A from the point $y=0$
l is the length that the car moves
$θ$ is the angle of the path of the car to the horizontal

Formula to calculate the potential energy of the system at point A is,

$PEA=mgyA$ (II)

m is mass of the car
g is acceleration due to gravity

Substitute equation (I) in (II),

$PEA=mg(lsinθ)$

Substitute $1.50×103 kg$ for $m$ , $9.8 m/s2$ for g, $50.0 m$ for l and $50.0°$ for $θ$ to find the potential energy of the system at point A,

$PEA=(1.50×103 kg)(9.8 m/s2)((50.0 m)sin50.0°)=5.63×105 J$

Thus, the potential energy of the system at point A is $5.63×105 J$ .

Section2:

To determine: The potential energy of the system at point B.

Answer: the potential energy of the system at point A is $1.69×105 J$ .

Explanation:

Given Info:

The mass of the roller-coaster car is $1.50×103 kg$ .

The distance that the car travelled between point B and C is $15.0 m$ .

The path is at an angle $50.0°$ below the horizontal to the lower point B.

Since $y=0$ is considered at point C,

$yB=lsinθ$ (I)

$yA=lsinθ$ (II)

$yB$ is the height to the point B from the point $y=0$
$yA$ is the height to the point A from the point $y=0$
l is the length that the car moves
$θ$ is the angle of the path of the car to the horizontal

Formula to calculate the potential energy of the system at point B is,

$PEB=mgyB$ (II)

m is mass of the car
g is acceleration due to gravity

Substitute equation (I) in (II),

$PEB=mg(lsinθ)$

Substitute $1.50×103 kg$ for $m$ , $9.8 m/s2$ for g, $15.0 m$ for l and $50.0°$ for $θ$ to find the potential energy of the system at point A,

$PEB=(1.50×103 kg)(9.8 m/s2)((15.0 m)sin50.0°)=1.69×105 J$

Thus, the potential energy of the system at point B is $1.69×105 J$ .

Conclusion:

From section1,

The potential energy of the system at point A is $5.63×105 J$ .

From section2,

The potential energy of the system on point B is $1.69×105 J$ .

Thus,

The change in potential energy when the system travels from A to B is,

$ΔPEA→B=PEB−PEA$

Substitute $1.69×105 J$ for $PEB$ and $5.63×105 J$ for $PEA$ to find the change in kinetic energy,

$ΔPEA→B=(1.69×105 J)−(5.63×105 J)=−3.94×105 J$

Therefore, the change in potential energy of the car when it moves from A to B is $−3.94×105 J$ when $y=0$ at point C which is $15.0 m$ down the slope from point B.

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Answer 2

Textbook Question

Chapter 5, Problem 63AP

A roller-coaster car of mass 1.50 × 10³ kg is initially at the top of a rise at point Chapter 5, Problem 63AP, A roller-coaster car of mass 1.50 103 kg is initially at the top of a rise at point . It then moves , example 1 . It then moves 35.0 m at an angle of 50.0° below the horizontal to a lower point . (a) Find both the potential energy of the system when the car is at points and Chapter 5, Problem 63AP, A roller-coaster car of mass 1.50 103 kg is initially at the top of a rise at point . It then moves , example 4 and the change in potential energy as the car moves from point to point , assuming y = 0 at point . (b) Repeat part (a), this time choosing y = 0 at point , which is another 15.0 m down the same slope from point Chapter 5, Problem 63AP, A roller-coaster car of mass 1.50 103 kg is initially at the top of a rise at point . It then moves , example 9 .

(a)

Expert Solution

To determine

Bothe the potential energy of the system when the car is on A and B and the change in potential energy of the car when it moves from A to B, when

$y=0$ at point B.

Answer to Problem 63AP

The potential energy of the system at point A is

$3.94×105 J$ .

The potential energy of the system on point B is zero.

The change in potential energy of the car when it moves from A to B is $−3.94×105 J$ when $y=0$ at point B.

Explanation of Solution

Section1:

To determine: The potential energy of the system at point A.

Answer: the potential energy of the system at point A is $3.94×105 J$ .

Explanation:

Given Info:

The mass of the roller-coaster car is $1.50×103 kg$ .

The distance that the car travelled between point A and B is $35.0 m$ .

The path is at an angle $50.0°$ below the horizontal to the lower point B.

Since $y=0$ is considered at point B,

$yB=0$

$yA=lsinθ$ (I)

$yB$ is the height to the point B from the point $y=0$
$yA$ is the height to the point A from the point $y=0$
l is the length that the car moves
$θ$ is the angle of the path of the car to the horizontal

Formula to calculate the potential energy of the system at point A is,

$PEA=mgyA$ (II)

m is mass of the car
g is acceleration due to gravity

Substitute equation (I) in (II),

$PEA=mg(lsinθ)$

Substitute $1.50×103 kg$ for $m$ , $9.8 m/s2$ for g, $35.0 m$ for l and $50.0°$ for $θ$ to find the potential energy of the system at point A,

$PEA=(1.50×103 kg)(9.8 m/s2)((35.0 m)sin50.0°)=3.94×105 J$

Thus, the potential energy of the system at point A is $3.94×105 J$ .

Section 2:

To determine: The potential energy of the system on point B

Answer: The potential energy of the system on point B is zero.

Explanation:

Since $y=0$ is considered at point B, the potential energy of the system at point B is zero.

Conclusion:

From section1,

The potential energy of the system at point A is $3.94×105 J$ .

From section2,

The potential energy of the system on point B is zero.

Thus,

The change in potential energy when the system travels from A to B is,

$ΔPEA→B=PEB−PEA$

Substitute zero for $PEB$ and $3.94×105 J$ for $PEA$ to find the change in kinetic energy,

$ΔPEA→B=0−3.94×105 J=−3.9×105 J$

Therefore, the change in potential energy of the car when it moves from A to B is $−3.9×105 J$ when $y=0$ at point B.

(b)

Expert Solution

To determine

Bothe the potential energy of the system when the car is on A and B and the change in potential energy of the car when it moves from A to B, when

$y=0$ at point C which is

$15.0 m$ down the slope from point B.

Answer to Problem 63AP

The potential energy of the system at point A is

$5.63×105 J$ .

The potential energy of the system on point B is $1.69×105 J$ .

The change in potential energy of the car when it moves from A to B is $−3.94×105 J$ when $y=0$ at point C which is $15.0 m$ down the slope from point B.

Explanation of Solution

Section1:

To determine: The potential energy of the system at point A.

Answer: the potential energy of the system at point A is $5.63×105 J$ .

Explanation:

Given Info:

The mass of the roller-coaster car is $1.50×103 kg$ .

The distance that the car travelled between point A and C is $50.0 m$ .

The path is at an angle $50.0°$ below the horizontal to the lower point B.

Since $y=0$ is considered at point C,

$yB=lsinθ$ (I)

$yA=lsinθ$ (II)

$yB$ is the height to the point B from the point $y=0$
$yA$ is the height to the point A from the point $y=0$
l is the length that the car moves
$θ$ is the angle of the path of the car to the horizontal

Formula to calculate the potential energy of the system at point A is,

$PEA=mgyA$ (II)

m is mass of the car
g is acceleration due to gravity

Substitute equation (I) in (II),

$PEA=mg(lsinθ)$

Substitute $1.50×103 kg$ for $m$ , $9.8 m/s2$ for g, $50.0 m$ for l and $50.0°$ for $θ$ to find the potential energy of the system at point A,

$PEA=(1.50×103 kg)(9.8 m/s2)((50.0 m)sin50.0°)=5.63×105 J$

Thus, the potential energy of the system at point A is $5.63×105 J$ .

Section2:

To determine: The potential energy of the system at point B.

Answer: the potential energy of the system at point A is $1.69×105 J$ .

Explanation:

Given Info:

The mass of the roller-coaster car is $1.50×103 kg$ .

The distance that the car travelled between point B and C is $15.0 m$ .

The path is at an angle $50.0°$ below the horizontal to the lower point B.

Since $y=0$ is considered at point C,

$yB=lsinθ$ (I)

$yA=lsinθ$ (II)

$yB$ is the height to the point B from the point $y=0$
$yA$ is the height to the point A from the point $y=0$
l is the length that the car moves
$θ$ is the angle of the path of the car to the horizontal

Formula to calculate the potential energy of the system at point B is,

$PEB=mgyB$ (II)

m is mass of the car
g is acceleration due to gravity

Substitute equation (I) in (II),

$PEB=mg(lsinθ)$

Substitute $1.50×103 kg$ for $m$ , $9.8 m/s2$ for g, $15.0 m$ for l and $50.0°$ for $θ$ to find the potential energy of the system at point A,

$PEB=(1.50×103 kg)(9.8 m/s2)((15.0 m)sin50.0°)=1.69×105 J$

Thus, the potential energy of the system at point B is $1.69×105 J$ .

Conclusion:

From section1,

The potential energy of the system at point A is $5.63×105 J$ .

From section2,

The potential energy of the system on point B is $1.69×105 J$ .

Thus,

The change in potential energy when the system travels from A to B is,

$ΔPEA→B=PEB−PEA$

Substitute $1.69×105 J$ for $PEB$ and $5.63×105 J$ for $PEA$ to find the change in kinetic energy,

$ΔPEA→B=(1.69×105 J)−(5.63×105 J)=−3.94×105 J$

Therefore, the change in potential energy of the car when it moves from A to B is $−3.94×105 J$ when $y=0$ at point C which is $15.0 m$ down the slope from point B.

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Answer 3

Textbook Question

Chapter 5, Problem 63AP

A roller-coaster car of mass 1.50 × 10³ kg is initially at the top of a rise at point Chapter 5, Problem 63AP, A roller-coaster car of mass 1.50 103 kg is initially at the top of a rise at point . It then moves , example 1 . It then moves 35.0 m at an angle of 50.0° below the horizontal to a lower point . (a) Find both the potential energy of the system when the car is at points and Chapter 5, Problem 63AP, A roller-coaster car of mass 1.50 103 kg is initially at the top of a rise at point . It then moves , example 4 and the change in potential energy as the car moves from point to point , assuming y = 0 at point . (b) Repeat part (a), this time choosing y = 0 at point , which is another 15.0 m down the same slope from point Chapter 5, Problem 63AP, A roller-coaster car of mass 1.50 103 kg is initially at the top of a rise at point . It then moves , example 9 .

(a)

Expert Solution

To determine

Bothe the potential energy of the system when the car is on A and B and the change in potential energy of the car when it moves from A to B, when

$y=0$ at point B.

Answer to Problem 63AP

The potential energy of the system at point A is

$3.94×105 J$ .

The potential energy of the system on point B is zero.

The change in potential energy of the car when it moves from A to B is $−3.94×105 J$ when $y=0$ at point B.

Explanation of Solution

Section1:

To determine: The potential energy of the system at point A.

Answer: the potential energy of the system at point A is $3.94×105 J$ .

Explanation:

Given Info:

The mass of the roller-coaster car is $1.50×103 kg$ .

The distance that the car travelled between point A and B is $35.0 m$ .

The path is at an angle $50.0°$ below the horizontal to the lower point B.

Since $y=0$ is considered at point B,

$yB=0$

$yA=lsinθ$ (I)

$yB$ is the height to the point B from the point $y=0$
$yA$ is the height to the point A from the point $y=0$
l is the length that the car moves
$θ$ is the angle of the path of the car to the horizontal

Formula to calculate the potential energy of the system at point A is,

$PEA=mgyA$ (II)

m is mass of the car
g is acceleration due to gravity

Substitute equation (I) in (II),

$PEA=mg(lsinθ)$

Substitute $1.50×103 kg$ for $m$ , $9.8 m/s2$ for g, $35.0 m$ for l and $50.0°$ for $θ$ to find the potential energy of the system at point A,

$PEA=(1.50×103 kg)(9.8 m/s2)((35.0 m)sin50.0°)=3.94×105 J$

Thus, the potential energy of the system at point A is $3.94×105 J$ .

Section 2:

To determine: The potential energy of the system on point B

Answer: The potential energy of the system on point B is zero.

Explanation:

Since $y=0$ is considered at point B, the potential energy of the system at point B is zero.

Conclusion:

From section1,

The potential energy of the system at point A is $3.94×105 J$ .

From section2,

The potential energy of the system on point B is zero.

Thus,

The change in potential energy when the system travels from A to B is,

$ΔPEA→B=PEB−PEA$

Substitute zero for $PEB$ and $3.94×105 J$ for $PEA$ to find the change in kinetic energy,

$ΔPEA→B=0−3.94×105 J=−3.9×105 J$

Therefore, the change in potential energy of the car when it moves from A to B is $−3.9×105 J$ when $y=0$ at point B.

(b)

Expert Solution

To determine

Bothe the potential energy of the system when the car is on A and B and the change in potential energy of the car when it moves from A to B, when

$y=0$ at point C which is

$15.0 m$ down the slope from point B.

Answer to Problem 63AP

The potential energy of the system at point A is

$5.63×105 J$ .

The potential energy of the system on point B is $1.69×105 J$ .

The change in potential energy of the car when it moves from A to B is $−3.94×105 J$ when $y=0$ at point C which is $15.0 m$ down the slope from point B.

Explanation of Solution

Section1:

To determine: The potential energy of the system at point A.

Answer: the potential energy of the system at point A is $5.63×105 J$ .

Explanation:

Given Info:

The mass of the roller-coaster car is $1.50×103 kg$ .

The distance that the car travelled between point A and C is $50.0 m$ .

The path is at an angle $50.0°$ below the horizontal to the lower point B.

Since $y=0$ is considered at point C,

$yB=lsinθ$ (I)

$yA=lsinθ$ (II)

$yB$ is the height to the point B from the point $y=0$
$yA$ is the height to the point A from the point $y=0$
l is the length that the car moves
$θ$ is the angle of the path of the car to the horizontal

Formula to calculate the potential energy of the system at point A is,

$PEA=mgyA$ (II)

m is mass of the car
g is acceleration due to gravity

Substitute equation (I) in (II),

$PEA=mg(lsinθ)$

Substitute $1.50×103 kg$ for $m$ , $9.8 m/s2$ for g, $50.0 m$ for l and $50.0°$ for $θ$ to find the potential energy of the system at point A,

$PEA=(1.50×103 kg)(9.8 m/s2)((50.0 m)sin50.0°)=5.63×105 J$

Thus, the potential energy of the system at point A is $5.63×105 J$ .

Section2:

To determine: The potential energy of the system at point B.

Answer: the potential energy of the system at point A is $1.69×105 J$ .

Explanation:

Given Info:

The mass of the roller-coaster car is $1.50×103 kg$ .

The distance that the car travelled between point B and C is $15.0 m$ .

The path is at an angle $50.0°$ below the horizontal to the lower point B.

Since $y=0$ is considered at point C,

$yB=lsinθ$ (I)

$yA=lsinθ$ (II)

$yB$ is the height to the point B from the point $y=0$
$yA$ is the height to the point A from the point $y=0$
l is the length that the car moves
$θ$ is the angle of the path of the car to the horizontal

Formula to calculate the potential energy of the system at point B is,

$PEB=mgyB$ (II)

m is mass of the car
g is acceleration due to gravity

Substitute equation (I) in (II),

$PEB=mg(lsinθ)$

Substitute $1.50×103 kg$ for $m$ , $9.8 m/s2$ for g, $15.0 m$ for l and $50.0°$ for $θ$ to find the potential energy of the system at point A,

$PEB=(1.50×103 kg)(9.8 m/s2)((15.0 m)sin50.0°)=1.69×105 J$

Thus, the potential energy of the system at point B is $1.69×105 J$ .

Conclusion:

From section1,

The potential energy of the system at point A is $5.63×105 J$ .

From section2,

The potential energy of the system on point B is $1.69×105 J$ .

Thus,

The change in potential energy when the system travels from A to B is,

$ΔPEA→B=PEB−PEA$

Substitute $1.69×105 J$ for $PEB$ and $5.63×105 J$ for $PEA$ to find the change in kinetic energy,

$ΔPEA→B=(1.69×105 J)−(5.63×105 J)=−3.94×105 J$

Therefore, the change in potential energy of the car when it moves from A to B is $−3.94×105 J$ when $y=0$ at point C which is $15.0 m$ down the slope from point B.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Textbook Question

Chapter 5, Problem 63AP

A roller-coaster car of mass 1.50 × 10³ kg is initially at the top of a rise at point Chapter 5, Problem 63AP, A roller-coaster car of mass 1.50 103 kg is initially at the top of a rise at point . It then moves , example 1 . It then moves 35.0 m at an angle of 50.0° below the horizontal to a lower point . (a) Find both the potential energy of the system when the car is at points and Chapter 5, Problem 63AP, A roller-coaster car of mass 1.50 103 kg is initially at the top of a rise at point . It then moves , example 4 and the change in potential energy as the car moves from point to point , assuming y = 0 at point . (b) Repeat part (a), this time choosing y = 0 at point , which is another 15.0 m down the same slope from point Chapter 5, Problem 63AP, A roller-coaster car of mass 1.50 103 kg is initially at the top of a rise at point . It then moves , example 9 .

(a)

Expert Solution

To determine

Bothe the potential energy of the system when the car is on A and B and the change in potential energy of the car when it moves from A to B, when

$y=0$ at point B.

Answer to Problem 63AP

The potential energy of the system at point A is

$3.94×105 J$ .

The potential energy of the system on point B is zero.

The change in potential energy of the car when it moves from A to B is $−3.94×105 J$ when $y=0$ at point B.

Explanation of Solution

Section1:

To determine: The potential energy of the system at point A.

Answer: the potential energy of the system at point A is $3.94×105 J$ .

Explanation:

Given Info:

The mass of the roller-coaster car is $1.50×103 kg$ .

The distance that the car travelled between point A and B is $35.0 m$ .

The path is at an angle $50.0°$ below the horizontal to the lower point B.

Since $y=0$ is considered at point B,

$yB=0$

$yA=lsinθ$ (I)

$yB$ is the height to the point B from the point $y=0$
$yA$ is the height to the point A from the point $y=0$
l is the length that the car moves
$θ$ is the angle of the path of the car to the horizontal

Formula to calculate the potential energy of the system at point A is,

$PEA=mgyA$ (II)

m is mass of the car
g is acceleration due to gravity

Substitute equation (I) in (II),

$PEA=mg(lsinθ)$

Substitute $1.50×103 kg$ for $m$ , $9.8 m/s2$ for g, $35.0 m$ for l and $50.0°$ for $θ$ to find the potential energy of the system at point A,

$PEA=(1.50×103 kg)(9.8 m/s2)((35.0 m)sin50.0°)=3.94×105 J$

Thus, the potential energy of the system at point A is $3.94×105 J$ .

Section 2:

To determine: The potential energy of the system on point B

Answer: The potential energy of the system on point B is zero.

Explanation:

Since $y=0$ is considered at point B, the potential energy of the system at point B is zero.

Conclusion:

From section1,

The potential energy of the system at point A is $3.94×105 J$ .

From section2,

The potential energy of the system on point B is zero.

Thus,

The change in potential energy when the system travels from A to B is,

$ΔPEA→B=PEB−PEA$

Substitute zero for $PEB$ and $3.94×105 J$ for $PEA$ to find the change in kinetic energy,

$ΔPEA→B=0−3.94×105 J=−3.9×105 J$

Therefore, the change in potential energy of the car when it moves from A to B is $−3.9×105 J$ when $y=0$ at point B.

(b)

Expert Solution

To determine

Bothe the potential energy of the system when the car is on A and B and the change in potential energy of the car when it moves from A to B, when

$y=0$ at point C which is

$15.0 m$ down the slope from point B.

Answer to Problem 63AP

The potential energy of the system at point A is

$5.63×105 J$ .

The potential energy of the system on point B is $1.69×105 J$ .

The change in potential energy of the car when it moves from A to B is $−3.94×105 J$ when $y=0$ at point C which is $15.0 m$ down the slope from point B.

Explanation of Solution

Section1:

To determine: The potential energy of the system at point A.

Answer: the potential energy of the system at point A is $5.63×105 J$ .

Explanation:

Given Info:

The mass of the roller-coaster car is $1.50×103 kg$ .

The distance that the car travelled between point A and C is $50.0 m$ .

The path is at an angle $50.0°$ below the horizontal to the lower point B.

Since $y=0$ is considered at point C,

$yB=lsinθ$ (I)

$yA=lsinθ$ (II)

$yB$ is the height to the point B from the point $y=0$
$yA$ is the height to the point A from the point $y=0$
l is the length that the car moves
$θ$ is the angle of the path of the car to the horizontal

Formula to calculate the potential energy of the system at point A is,

$PEA=mgyA$ (II)

m is mass of the car
g is acceleration due to gravity

Substitute equation (I) in (II),

$PEA=mg(lsinθ)$

Substitute $1.50×103 kg$ for $m$ , $9.8 m/s2$ for g, $50.0 m$ for l and $50.0°$ for $θ$ to find the potential energy of the system at point A,

$PEA=(1.50×103 kg)(9.8 m/s2)((50.0 m)sin50.0°)=5.63×105 J$

Thus, the potential energy of the system at point A is $5.63×105 J$ .

Section2:

To determine: The potential energy of the system at point B.

Answer: the potential energy of the system at point A is $1.69×105 J$ .

Explanation:

Given Info:

The mass of the roller-coaster car is $1.50×103 kg$ .

The distance that the car travelled between point B and C is $15.0 m$ .

The path is at an angle $50.0°$ below the horizontal to the lower point B.

Since $y=0$ is considered at point C,

$yB=lsinθ$ (I)

$yA=lsinθ$ (II)

$yB$ is the height to the point B from the point $y=0$
$yA$ is the height to the point A from the point $y=0$
l is the length that the car moves
$θ$ is the angle of the path of the car to the horizontal

Formula to calculate the potential energy of the system at point B is,

$PEB=mgyB$ (II)

m is mass of the car
g is acceleration due to gravity

Substitute equation (I) in (II),

$PEB=mg(lsinθ)$

Substitute $1.50×103 kg$ for $m$ , $9.8 m/s2$ for g, $15.0 m$ for l and $50.0°$ for $θ$ to find the potential energy of the system at point A,

$PEB=(1.50×103 kg)(9.8 m/s2)((15.0 m)sin50.0°)=1.69×105 J$

Thus, the potential energy of the system at point B is $1.69×105 J$ .

Conclusion:

From section1,

The potential energy of the system at point A is $5.63×105 J$ .

From section2,

The potential energy of the system on point B is $1.69×105 J$ .

Thus,

The change in potential energy when the system travels from A to B is,

$ΔPEA→B=PEB−PEA$

Substitute $1.69×105 J$ for $PEB$ and $5.63×105 J$ for $PEA$ to find the change in kinetic energy,

$ΔPEA→B=(1.69×105 J)−(5.63×105 J)=−3.94×105 J$

Therefore, the change in potential energy of the car when it moves from A to B is $−3.94×105 J$ when $y=0$ at point C which is $15.0 m$ down the slope from point B.

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

To determine

Bothe the potential energy of the system when the car is on A and B and the change in potential energy of the car when it moves from A to B, when

$y=0$ at point B.

Answer to Problem 63AP

The potential energy of the system at point A is

$3.94×105 J$ .

The potential energy of the system on point B is zero.

The change in potential energy of the car when it moves from A to B is $−3.94×105 J$ when $y=0$ at point B.

Explanation of Solution

Section1:

To determine: The potential energy of the system at point A.

Answer: the potential energy of the system at point A is $3.94×105 J$ .

Explanation:

Given Info:

The mass of the roller-coaster car is $1.50×103 kg$ .

The distance that the car travelled between point A and B is $35.0 m$ .

The path is at an angle $50.0°$ below the horizontal to the lower point B.

Since $y=0$ is considered at point B,

$yB=0$

$yA=lsinθ$ (I)

$yB$ is the height to the point B from the point $y=0$
$yA$ is the height to the point A from the point $y=0$
l is the length that the car moves
$θ$ is the angle of the path of the car to the horizontal

Formula to calculate the potential energy of the system at point A is,

$PEA=mgyA$ (II)

m is mass of the car
g is acceleration due to gravity

Substitute equation (I) in (II),

$PEA=mg(lsinθ)$

Substitute $1.50×103 kg$ for $m$ , $9.8 m/s2$ for g, $35.0 m$ for l and $50.0°$ for $θ$ to find the potential energy of the system at point A,

$PEA=(1.50×103 kg)(9.8 m/s2)((35.0 m)sin50.0°)=3.94×105 J$

Thus, the potential energy of the system at point A is $3.94×105 J$ .

Section 2:

To determine: The potential energy of the system on point B

Answer: The potential energy of the system on point B is zero.

Explanation:

Since $y=0$ is considered at point B, the potential energy of the system at point B is zero.

Conclusion:

From section1,

The potential energy of the system at point A is $3.94×105 J$ .

From section2,

The potential energy of the system on point B is zero.

Thus,

The change in potential energy when the system travels from A to B is,

$ΔPEA→B=PEB−PEA$

Substitute zero for $PEB$ and $3.94×105 J$ for $PEA$ to find the change in kinetic energy,

$ΔPEA→B=0−3.94×105 J=−3.9×105 J$

Therefore, the change in potential energy of the car when it moves from A to B is $−3.9×105 J$ when $y=0$ at point B.

(b)

Expert Solution

To determine

Bothe the potential energy of the system when the car is on A and B and the change in potential energy of the car when it moves from A to B, when

$y=0$ at point C which is

$15.0 m$ down the slope from point B.

Answer to Problem 63AP

The potential energy of the system at point A is

$5.63×105 J$ .

The potential energy of the system on point B is $1.69×105 J$ .

The change in potential energy of the car when it moves from A to B is $−3.94×105 J$ when $y=0$ at point C which is $15.0 m$ down the slope from point B.

Explanation of Solution

Section1:

To determine: The potential energy of the system at point A.

Answer: the potential energy of the system at point A is $5.63×105 J$ .

Explanation:

Given Info:

The mass of the roller-coaster car is $1.50×103 kg$ .

The distance that the car travelled between point A and C is $50.0 m$ .

The path is at an angle $50.0°$ below the horizontal to the lower point B.

Since $y=0$ is considered at point C,

$yB=lsinθ$ (I)

$yA=lsinθ$ (II)

$yB$ is the height to the point B from the point $y=0$
$yA$ is the height to the point A from the point $y=0$
l is the length that the car moves
$θ$ is the angle of the path of the car to the horizontal

Formula to calculate the potential energy of the system at point A is,

$PEA=mgyA$ (II)

m is mass of the car
g is acceleration due to gravity

Substitute equation (I) in (II),

$PEA=mg(lsinθ)$

Substitute $1.50×103 kg$ for $m$ , $9.8 m/s2$ for g, $50.0 m$ for l and $50.0°$ for $θ$ to find the potential energy of the system at point A,

$PEA=(1.50×103 kg)(9.8 m/s2)((50.0 m)sin50.0°)=5.63×105 J$

Thus, the potential energy of the system at point A is $5.63×105 J$ .

Section2:

To determine: The potential energy of the system at point B.

Answer: the potential energy of the system at point A is $1.69×105 J$ .

Explanation:

Given Info:

The mass of the roller-coaster car is $1.50×103 kg$ .

The distance that the car travelled between point B and C is $15.0 m$ .

The path is at an angle $50.0°$ below the horizontal to the lower point B.

Since $y=0$ is considered at point C,

$yB=lsinθ$ (I)

$yA=lsinθ$ (II)

$yB$ is the height to the point B from the point $y=0$
$yA$ is the height to the point A from the point $y=0$
l is the length that the car moves
$θ$ is the angle of the path of the car to the horizontal

Formula to calculate the potential energy of the system at point B is,

$PEB=mgyB$ (II)

m is mass of the car
g is acceleration due to gravity

Substitute equation (I) in (II),

$PEB=mg(lsinθ)$

Substitute $1.50×103 kg$ for $m$ , $9.8 m/s2$ for g, $15.0 m$ for l and $50.0°$ for $θ$ to find the potential energy of the system at point A,

$PEB=(1.50×103 kg)(9.8 m/s2)((15.0 m)sin50.0°)=1.69×105 J$

Thus, the potential energy of the system at point B is $1.69×105 J$ .

Conclusion:

From section1,

The potential energy of the system at point A is $5.63×105 J$ .

From section2,

The potential energy of the system on point B is $1.69×105 J$ .

Thus,

The change in potential energy when the system travels from A to B is,

$ΔPEA→B=PEB−PEA$

Substitute $1.69×105 J$ for $PEB$ and $5.63×105 J$ for $PEA$ to find the change in kinetic energy,

$ΔPEA→B=(1.69×105 J)−(5.63×105 J)=−3.94×105 J$

Therefore, the change in potential energy of the car when it moves from A to B is $−3.94×105 J$ when $y=0$ at point C which is $15.0 m$ down the slope from point B.

Answer 8

(a)

Expert Solution

To determine

Bothe the potential energy of the system when the car is on A and B and the change in potential energy of the car when it moves from A to B, when

$y=0$ at point B.

Answer to Problem 63AP

The potential energy of the system at point A is

$3.94×105 J$ .

The potential energy of the system on point B is zero.

The change in potential energy of the car when it moves from A to B is $−3.94×105 J$ when $y=0$ at point B.

Explanation of Solution

Section1:

To determine: The potential energy of the system at point A.

Answer: the potential energy of the system at point A is $3.94×105 J$ .

Explanation:

Given Info:

The mass of the roller-coaster car is $1.50×103 kg$ .

The distance that the car travelled between point A and B is $35.0 m$ .

The path is at an angle $50.0°$ below the horizontal to the lower point B.

Since $y=0$ is considered at point B,

$yB=0$

$yA=lsinθ$ (I)

$yB$ is the height to the point B from the point $y=0$
$yA$ is the height to the point A from the point $y=0$
l is the length that the car moves
$θ$ is the angle of the path of the car to the horizontal

Formula to calculate the potential energy of the system at point A is,

$PEA=mgyA$ (II)

m is mass of the car
g is acceleration due to gravity

Substitute equation (I) in (II),

$PEA=mg(lsinθ)$

Substitute $1.50×103 kg$ for $m$ , $9.8 m/s2$ for g, $35.0 m$ for l and $50.0°$ for $θ$ to find the potential energy of the system at point A,

$PEA=(1.50×103 kg)(9.8 m/s2)((35.0 m)sin50.0°)=3.94×105 J$

Thus, the potential energy of the system at point A is $3.94×105 J$ .

Section 2:

To determine: The potential energy of the system on point B

Answer: The potential energy of the system on point B is zero.

Explanation:

Since $y=0$ is considered at point B, the potential energy of the system at point B is zero.

Conclusion:

From section1,

The potential energy of the system at point A is $3.94×105 J$ .

From section2,

The potential energy of the system on point B is zero.

Thus,

The change in potential energy when the system travels from A to B is,

$ΔPEA→B=PEB−PEA$

Substitute zero for $PEB$ and $3.94×105 J$ for $PEA$ to find the change in kinetic energy,

$ΔPEA→B=0−3.94×105 J=−3.9×105 J$

Therefore, the change in potential energy of the car when it moves from A to B is $−3.9×105 J$ when $y=0$ at point B.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

Bothe the potential energy of the system when the car is on A and B and the change in potential energy of the car when it moves from A to B, when

$y=0$ at point B.

Answer 13

Answer to Problem 63AP

The potential energy of the system at point A is

$3.94×105 J$ .

The potential energy of the system on point B is zero.

The change in potential energy of the car when it moves from A to B is $−3.94×105 J$ when $y=0$ at point B.

Answer 14

Explanation of Solution

Section1:

To determine: The potential energy of the system at point A.

Answer: the potential energy of the system at point A is $3.94×105 J$ .

Explanation:

Given Info:

The mass of the roller-coaster car is $1.50×103 kg$ .

The distance that the car travelled between point A and B is $35.0 m$ .

The path is at an angle $50.0°$ below the horizontal to the lower point B.

Since $y=0$ is considered at point B,

$yB=0$

$yA=lsinθ$ (I)

$yB$ is the height to the point B from the point $y=0$
$yA$ is the height to the point A from the point $y=0$
l is the length that the car moves
$θ$ is the angle of the path of the car to the horizontal

Formula to calculate the potential energy of the system at point A is,

$PEA=mgyA$ (II)

m is mass of the car
g is acceleration due to gravity

Substitute equation (I) in (II),

$PEA=mg(lsinθ)$

Substitute $1.50×103 kg$ for $m$ , $9.8 m/s2$ for g, $35.0 m$ for l and $50.0°$ for $θ$ to find the potential energy of the system at point A,

$PEA=(1.50×103 kg)(9.8 m/s2)((35.0 m)sin50.0°)=3.94×105 J$

Thus, the potential energy of the system at point A is $3.94×105 J$ .

Section 2:

To determine: The potential energy of the system on point B

Answer: The potential energy of the system on point B is zero.

Explanation:

Since $y=0$ is considered at point B, the potential energy of the system at point B is zero.

Conclusion:

From section1,

The potential energy of the system at point A is $3.94×105 J$ .

From section2,

The potential energy of the system on point B is zero.

Thus,

The change in potential energy when the system travels from A to B is,

$ΔPEA→B=PEB−PEA$

Substitute zero for $PEB$ and $3.94×105 J$ for $PEA$ to find the change in kinetic energy,

$ΔPEA→B=0−3.94×105 J=−3.9×105 J$

Therefore, the change in potential energy of the car when it moves from A to B is $−3.9×105 J$ when $y=0$ at point B.

Answer 15

(b)

Expert Solution

To determine

Bothe the potential energy of the system when the car is on A and B and the change in potential energy of the car when it moves from A to B, when

$y=0$ at point C which is

$15.0 m$ down the slope from point B.

Answer to Problem 63AP

The potential energy of the system at point A is

$5.63×105 J$ .

The potential energy of the system on point B is $1.69×105 J$ .

The change in potential energy of the car when it moves from A to B is $−3.94×105 J$ when $y=0$ at point C which is $15.0 m$ down the slope from point B.

Explanation of Solution

Section1:

To determine: The potential energy of the system at point A.

Answer: the potential energy of the system at point A is $5.63×105 J$ .

Explanation:

Given Info:

The mass of the roller-coaster car is $1.50×103 kg$ .

The distance that the car travelled between point A and C is $50.0 m$ .

The path is at an angle $50.0°$ below the horizontal to the lower point B.

Since $y=0$ is considered at point C,

$yB=lsinθ$ (I)

$yA=lsinθ$ (II)

$yB$ is the height to the point B from the point $y=0$
$yA$ is the height to the point A from the point $y=0$
l is the length that the car moves
$θ$ is the angle of the path of the car to the horizontal

Formula to calculate the potential energy of the system at point A is,

$PEA=mgyA$ (II)

m is mass of the car
g is acceleration due to gravity

Substitute equation (I) in (II),

$PEA=mg(lsinθ)$

Substitute $1.50×103 kg$ for $m$ , $9.8 m/s2$ for g, $50.0 m$ for l and $50.0°$ for $θ$ to find the potential energy of the system at point A,

$PEA=(1.50×103 kg)(9.8 m/s2)((50.0 m)sin50.0°)=5.63×105 J$

Thus, the potential energy of the system at point A is $5.63×105 J$ .

Section2:

To determine: The potential energy of the system at point B.

Answer: the potential energy of the system at point A is $1.69×105 J$ .

Explanation:

Given Info:

The mass of the roller-coaster car is $1.50×103 kg$ .

The distance that the car travelled between point B and C is $15.0 m$ .

The path is at an angle $50.0°$ below the horizontal to the lower point B.

Since $y=0$ is considered at point C,

$yB=lsinθ$ (I)

$yA=lsinθ$ (II)

$yB$ is the height to the point B from the point $y=0$
$yA$ is the height to the point A from the point $y=0$
l is the length that the car moves
$θ$ is the angle of the path of the car to the horizontal

Formula to calculate the potential energy of the system at point B is,

$PEB=mgyB$ (II)

m is mass of the car
g is acceleration due to gravity

Substitute equation (I) in (II),

$PEB=mg(lsinθ)$

Substitute $1.50×103 kg$ for $m$ , $9.8 m/s2$ for g, $15.0 m$ for l and $50.0°$ for $θ$ to find the potential energy of the system at point A,

$PEB=(1.50×103 kg)(9.8 m/s2)((15.0 m)sin50.0°)=1.69×105 J$

Thus, the potential energy of the system at point B is $1.69×105 J$ .

Conclusion:

From section1,

The potential energy of the system at point A is $5.63×105 J$ .

From section2,

The potential energy of the system on point B is $1.69×105 J$ .

Thus,

The change in potential energy when the system travels from A to B is,

$ΔPEA→B=PEB−PEA$

Substitute $1.69×105 J$ for $PEB$ and $5.63×105 J$ for $PEA$ to find the change in kinetic energy,

$ΔPEA→B=(1.69×105 J)−(5.63×105 J)=−3.94×105 J$

Therefore, the change in potential energy of the car when it moves from A to B is $−3.94×105 J$ when $y=0$ at point C which is $15.0 m$ down the slope from point B.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

Bothe the potential energy of the system when the car is on A and B and the change in potential energy of the car when it moves from A to B, when

$y=0$ at point C which is

$15.0 m$ down the slope from point B.

Answer 20

Answer to Problem 63AP

The potential energy of the system at point A is

$5.63×105 J$ .

The potential energy of the system on point B is $1.69×105 J$ .

The change in potential energy of the car when it moves from A to B is $−3.94×105 J$ when $y=0$ at point C which is $15.0 m$ down the slope from point B.

Answer 21

Explanation of Solution

Section1:

To determine: The potential energy of the system at point A.

Answer: the potential energy of the system at point A is $5.63×105 J$ .

Explanation:

Given Info:

The mass of the roller-coaster car is $1.50×103 kg$ .

The distance that the car travelled between point A and C is $50.0 m$ .

The path is at an angle $50.0°$ below the horizontal to the lower point B.

Since $y=0$ is considered at point C,

$yB=lsinθ$ (I)

$yA=lsinθ$ (II)

$yB$ is the height to the point B from the point $y=0$
$yA$ is the height to the point A from the point $y=0$
l is the length that the car moves
$θ$ is the angle of the path of the car to the horizontal

Formula to calculate the potential energy of the system at point A is,

$PEA=mgyA$ (II)

m is mass of the car
g is acceleration due to gravity

Substitute equation (I) in (II),

$PEA=mg(lsinθ)$

Substitute $1.50×103 kg$ for $m$ , $9.8 m/s2$ for g, $50.0 m$ for l and $50.0°$ for $θ$ to find the potential energy of the system at point A,

$PEA=(1.50×103 kg)(9.8 m/s2)((50.0 m)sin50.0°)=5.63×105 J$

Thus, the potential energy of the system at point A is $5.63×105 J$ .

Section2:

To determine: The potential energy of the system at point B.

Answer: the potential energy of the system at point A is $1.69×105 J$ .

Explanation:

Given Info:

The mass of the roller-coaster car is $1.50×103 kg$ .

The distance that the car travelled between point B and C is $15.0 m$ .

The path is at an angle $50.0°$ below the horizontal to the lower point B.

Since $y=0$ is considered at point C,

$yB=lsinθ$ (I)

$yA=lsinθ$ (II)

$yB$ is the height to the point B from the point $y=0$
$yA$ is the height to the point A from the point $y=0$
l is the length that the car moves
$θ$ is the angle of the path of the car to the horizontal

Formula to calculate the potential energy of the system at point B is,

$PEB=mgyB$ (II)

m is mass of the car
g is acceleration due to gravity

Substitute equation (I) in (II),

$PEB=mg(lsinθ)$

Substitute $1.50×103 kg$ for $m$ , $9.8 m/s2$ for g, $15.0 m$ for l and $50.0°$ for $θ$ to find the potential energy of the system at point A,

$PEB=(1.50×103 kg)(9.8 m/s2)((15.0 m)sin50.0°)=1.69×105 J$

Thus, the potential energy of the system at point B is $1.69×105 J$ .

Conclusion:

From section1,

The potential energy of the system at point A is $5.63×105 J$ .

From section2,

The potential energy of the system on point B is $1.69×105 J$ .

Thus,

The change in potential energy when the system travels from A to B is,

$ΔPEA→B=PEB−PEA$

Substitute $1.69×105 J$ for $PEB$ and $5.63×105 J$ for $PEA$ to find the change in kinetic energy,

$ΔPEA→B=(1.69×105 J)−(5.63×105 J)=−3.94×105 J$

Therefore, the change in potential energy of the car when it moves from A to B is $−3.94×105 J$ when $y=0$ at point C which is $15.0 m$ down the slope from point B.