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 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 .

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 $3.94\times {10}^5\text{J}$.

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 $-3.94\times {10}^5\text{J}$ 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 $3.94\times {10}^5\text{J}$.

Explanation:

Given Info:

The mass of the roller-coaster car is $1.50\times {10}^3\text{kg}$.

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

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

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

$y_B=0$

$y_A=l\sin \theta$      (I)

• $y_B$ is the height to the point B from the point $y=0$
• $y_A$ is the height to the point A from the point $y=0$
• l is the length that the car moves
• $\theta$ 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,

$P{E}_A=mg{y}_A$(II)

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

Substitute equation (I) in (II),

$P{E}_A=mg\left(l\sin \theta \right)$

Substitute $1.50\times {10}^3\text{kg}$ for $m$, $9.8{\text{m/s}}^{\text{2}}$ for g, $35.0\text{m}$ for l and $50.0°$ for $\theta$ to find the potential energy of the system at point A,

$\begin{array}{c}P{E}_A=\left(1.50\times {10}^3\text{kg}\right)\left(9.8{\text{m/s}}^{\text{2}}\right)\left(\left(35.0\text{m}\right)\sin 50.0°\right)\\ =3.94\times {10}^5\text{J}\end{array}$

Thus, the potential energy of the system at point A is $3.94\times {10}^5\text{J}$.

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 $3.94\times {10}^5\text{J}$.

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,

$\Delta P{E}_{A\to B}=P{E}_B-P{E}_A$

Substitute zero for $P{E}_B$ and $3.94\times {10}^5\text{J}$ for $P{E}_A$ to find the change in kinetic energy,

$\begin{array}{c}\Delta P{E}_{A\to B}=0-3.94\times {10}^5\text{J}\\ =-3.9\times {10}^5\text{J}\end{array}$

Therefore, the change in potential energy of the car when it moves from A to B is $-3.9\times {10}^5\text{J}$ when $y=0$ at point B.

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 $15.0\text{m}$ down the slope from point B.

Answer to Problem 63AP

The potential energy of the system at point A is $5.63\times {10}^5\text{J}$.

The potential energy of the system on point B is $1.69\times {10}^5\text{J}$.

The change in potential energy of the car when it moves from A to B is $-3.94\times {10}^5\text{J}$ when $y=0$ at point C which is $15.0\text{m}$ 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 $5.63\times {10}^5\text{J}$.

Explanation:

Given Info:

The mass of the roller-coaster car is $1.50\times {10}^3\text{kg}$.

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

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

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

$y_B=l\sin \theta$      (I)

$y_A=l\sin \theta$      (II)

• $y_B$ is the height to the point B from the point $y=0$
• $y_A$ is the height to the point A from the point $y=0$
• l is the length that the car moves
• $\theta$ 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,

$P{E}_A=mg{y}_A$(II)

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

Substitute equation (I) in (II),

$P{E}_A=mg\left(l\sin \theta \right)$

Substitute $1.50\times {10}^3\text{kg}$ for $m$, $9.8{\text{m/s}}^{\text{2}}$ for g, $50.0\text{m}$ for l and $50.0°$ for $\theta$ to find the potential energy of the system at point A,

$\begin{array}{c}P{E}_A=\left(1.50\times {10}^3\text{kg}\right)\left(9.8{\text{m/s}}^{\text{2}}\right)\left(\left(50.0\text{m}\right)\sin 50.0°\right)\\ =5.63\times {10}^5\text{J}\end{array}$

Thus, the potential energy of the system at point A is $5.63\times {10}^5\text{J}$.

Section2:

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

Answer: the potential energy of the system at point A is $1.69\times {10}^5\text{J}$.

Explanation:

Given Info:

The mass of the roller-coaster car is $1.50\times {10}^3\text{kg}$.

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

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

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

$y_B=l\sin \theta$      (I)

$y_A=l\sin \theta$      (II)

• $y_B$ is the height to the point B from the point $y=0$
• $y_A$ is the height to the point A from the point $y=0$
• l is the length that the car moves
• $\theta$ 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,

$P{E}_B=mg{y}_B$(II)

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

Substitute equation (I) in (II),

$P{E}_B=mg\left(l\sin \theta \right)$

Substitute $1.50\times {10}^3\text{kg}$ for $m$, $9.8{\text{m/s}}^{\text{2}}$ for g, $15.0\text{m}$ for l and $50.0°$ for $\theta$ to find the potential energy of the system at point A,

$\begin{array}{c}P{E}_B=\left(1.50\times {10}^3\text{kg}\right)\left(9.8{\text{m/s}}^{\text{2}}\right)\left(\left(15.0\text{m}\right)\sin 50.0°\right)\\ =1.69\times {10}^5\text{J}\end{array}$

Thus, the potential energy of the system at point B is $1.69\times {10}^5\text{J}$.

Conclusion:

From section1,

The potential energy of the system at point A is $5.63\times {10}^5\text{J}$.

From section2,

The potential energy of the system on point B is $1.69\times {10}^5\text{J}$.

Thus,

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

$\Delta P{E}_{A\to B}=P{E}_B-P{E}_A$

Substitute $1.69\times {10}^5\text{J}$ for $P{E}_B$ and $5.63\times {10}^5\text{J}$ for $P{E}_A$ to find the change in kinetic energy,

$\begin{array}{c}\Delta P{E}_{A\to B}=\left(1.69\times {10}^5\text{J}\right)-\left(5.63\times {10}^5\text{J}\right)\\ =-3.94\times {10}^5\text{J}\end{array}$

Therefore, the change in potential energy of the car when it moves from A to B is $-3.94\times {10}^5\text{J}$ when $y=0$ at point C which is $15.0\text{m}$ down the slope from point B.