The amount of energy dissipated by friction.

Question

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Answer 1

Question

Chapter 6, Problem 100P

(a)

To determine

The amount of energy dissipated by friction.

(a)

Expert Solution

Answer to Problem 100P

The amount of energy dissipated by friction is $935 J_$ .

Explanation of Solution

Figure 1 represents the free body diagram of the crate.

COLLEGE PHYICS, Chapter 6, Problem 100P

For an isolated system the total energy of the system (the sum of mechanical energy and its internal energy) is conserved.

$ΔK+ΔU+ΔEint=0$

Here, $ΔK$ is the change in kinetic energy, $ΔU$ is the change in potential energy, and $ΔEint$ is the change in internal energy of the system.

In this situation, the work done by the frictional force is equal to the change in internal energy of the system.

The above equation is reduced to.

$(12mvf2−12mvi2)+mgΔy+ΔEint=0$

Here, $ΔEint$ is the energy dissipated by friction, $m$ is the mass of the crate, $g$ is the acceleration due to gravity, and $Δy$ is the distance travelled along the incline.

Since the final velocity of the crate is zero and the distance travelled along incline is given by $−dsinθ$ , the above equation is reduced to

$−12mvi2−mgdsinθ+ΔEint=0$ (I)

Here, $d$ is the distance travelled along the incline.

Rearrange the above equation.

$ΔEint=12mvi2+mgdsinθ$ (II)

Conclusion:

Substitute $100 kg$ for $m$ , $2 m/s$ for $vi$ , $9.8 m/s2$ for $g$ , $1.5 m$ for $d$ , and $30°$ for $θ$ in equation (I), to find $ΔEint$ .

$ΔEint=12(100 kg)(2 m/s)2−(100 kg)(9.8 m/s2)(1.5 m)sin30°=935 J$

Therefore, the amount of energy dissipated by friction is $935 J_$ .

(b)

To determine

The coefficient of the sliding friction.

(b)

Expert Solution

Answer to Problem 100P

The coefficient of the sliding friction is $0.73_$ .

Explanation of Solution

Write the expression for net force acting along y direction.

$∑Fy=0$

Apply the above condition in Figure 1.

$N−mgcosθ=0N=mgcosθ$ (III)

Write the expression for internal energy dissipated by friction.

$ΔEfric=μkNd$

Here, $μk$ is the coefficient of kinetic friction, $N$ is the normal force, and $d$ is the distance travelled along the incline.

Use the above equation in equation (III).

$ΔEfric=μkmgcosθd$

Rearrange the above equation to find $μk$ .

$μk=ΔEfricmgdcosθ$ (IV)

Conclusion:

Substitute $935 J$ for $ΔEint$ , $100 kg$ for $m$ , $9.8 m/s2$ for $g$ , $1.5 m$ for $d$ , and $30°$ for $θ$ in equation (IV), to find $μk$ .

$μk=935 J(100 kg)(9.8 m/s2)(1.5 m)cos30°=0.73$

Therefore, the coefficient of the sliding friction is $0.73_$ .

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A man slides two boxes up a slope. The two boxes A and B have a mass of 75 kg and 50 kg, respectively. (a) Draw the free body diagram (FBD) of the two crates. (b) Determine the tension in the cable that the man must exert to cause imminent movement from rest of the two boxes. Static friction coefficient USA = 0.25 HSB = 0.35 Kinetic friction coefficient HkA = 0.20 HkB = 0.25 M₁ = 75 kg MB = 50 kg P 35° Figure 3 B 200

Answer 2

Question

Chapter 6, Problem 100P

(a)

To determine

The amount of energy dissipated by friction.

(a)

Expert Solution

Answer to Problem 100P

The amount of energy dissipated by friction is $935 J_$ .

Explanation of Solution

Figure 1 represents the free body diagram of the crate.

COLLEGE PHYICS, Chapter 6, Problem 100P

For an isolated system the total energy of the system (the sum of mechanical energy and its internal energy) is conserved.

$ΔK+ΔU+ΔEint=0$

Here, $ΔK$ is the change in kinetic energy, $ΔU$ is the change in potential energy, and $ΔEint$ is the change in internal energy of the system.

In this situation, the work done by the frictional force is equal to the change in internal energy of the system.

The above equation is reduced to.

$(12mvf2−12mvi2)+mgΔy+ΔEint=0$

Here, $ΔEint$ is the energy dissipated by friction, $m$ is the mass of the crate, $g$ is the acceleration due to gravity, and $Δy$ is the distance travelled along the incline.

Since the final velocity of the crate is zero and the distance travelled along incline is given by $−dsinθ$ , the above equation is reduced to

$−12mvi2−mgdsinθ+ΔEint=0$ (I)

Here, $d$ is the distance travelled along the incline.

Rearrange the above equation.

$ΔEint=12mvi2+mgdsinθ$ (II)

Conclusion:

Substitute $100 kg$ for $m$ , $2 m/s$ for $vi$ , $9.8 m/s2$ for $g$ , $1.5 m$ for $d$ , and $30°$ for $θ$ in equation (I), to find $ΔEint$ .

$ΔEint=12(100 kg)(2 m/s)2−(100 kg)(9.8 m/s2)(1.5 m)sin30°=935 J$

Therefore, the amount of energy dissipated by friction is $935 J_$ .

(b)

To determine

The coefficient of the sliding friction.

(b)

Expert Solution

Answer to Problem 100P

The coefficient of the sliding friction is $0.73_$ .

Explanation of Solution

Write the expression for net force acting along y direction.

$∑Fy=0$

Apply the above condition in Figure 1.

$N−mgcosθ=0N=mgcosθ$ (III)

Write the expression for internal energy dissipated by friction.

$ΔEfric=μkNd$

Here, $μk$ is the coefficient of kinetic friction, $N$ is the normal force, and $d$ is the distance travelled along the incline.

Use the above equation in equation (III).

$ΔEfric=μkmgcosθd$

Rearrange the above equation to find $μk$ .

$μk=ΔEfricmgdcosθ$ (IV)

Conclusion:

Substitute $935 J$ for $ΔEint$ , $100 kg$ for $m$ , $9.8 m/s2$ for $g$ , $1.5 m$ for $d$ , and $30°$ for $θ$ in equation (IV), to find $μk$ .

$μk=935 J(100 kg)(9.8 m/s2)(1.5 m)cos30°=0.73$

Therefore, the coefficient of the sliding friction is $0.73_$ .

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Answer 3

Question

Chapter 6, Problem 100P

(a)

To determine

The amount of energy dissipated by friction.

(a)

Expert Solution

Answer to Problem 100P

The amount of energy dissipated by friction is $935 J_$ .

Explanation of Solution

Figure 1 represents the free body diagram of the crate.

COLLEGE PHYICS, Chapter 6, Problem 100P

For an isolated system the total energy of the system (the sum of mechanical energy and its internal energy) is conserved.

$ΔK+ΔU+ΔEint=0$

Here, $ΔK$ is the change in kinetic energy, $ΔU$ is the change in potential energy, and $ΔEint$ is the change in internal energy of the system.

In this situation, the work done by the frictional force is equal to the change in internal energy of the system.

The above equation is reduced to.

$(12mvf2−12mvi2)+mgΔy+ΔEint=0$

Here, $ΔEint$ is the energy dissipated by friction, $m$ is the mass of the crate, $g$ is the acceleration due to gravity, and $Δy$ is the distance travelled along the incline.

Since the final velocity of the crate is zero and the distance travelled along incline is given by $−dsinθ$ , the above equation is reduced to

$−12mvi2−mgdsinθ+ΔEint=0$ (I)

Here, $d$ is the distance travelled along the incline.

Rearrange the above equation.

$ΔEint=12mvi2+mgdsinθ$ (II)

Conclusion:

Substitute $100 kg$ for $m$ , $2 m/s$ for $vi$ , $9.8 m/s2$ for $g$ , $1.5 m$ for $d$ , and $30°$ for $θ$ in equation (I), to find $ΔEint$ .

$ΔEint=12(100 kg)(2 m/s)2−(100 kg)(9.8 m/s2)(1.5 m)sin30°=935 J$

Therefore, the amount of energy dissipated by friction is $935 J_$ .

(b)

To determine

The coefficient of the sliding friction.

(b)

Expert Solution

Answer to Problem 100P

The coefficient of the sliding friction is $0.73_$ .

Explanation of Solution

Write the expression for net force acting along y direction.

$∑Fy=0$

Apply the above condition in Figure 1.

$N−mgcosθ=0N=mgcosθ$ (III)

Write the expression for internal energy dissipated by friction.

$ΔEfric=μkNd$

Here, $μk$ is the coefficient of kinetic friction, $N$ is the normal force, and $d$ is the distance travelled along the incline.

Use the above equation in equation (III).

$ΔEfric=μkmgcosθd$

Rearrange the above equation to find $μk$ .

$μk=ΔEfricmgdcosθ$ (IV)

Conclusion:

Substitute $935 J$ for $ΔEint$ , $100 kg$ for $m$ , $9.8 m/s2$ for $g$ , $1.5 m$ for $d$ , and $30°$ for $θ$ in equation (IV), to find $μk$ .

$μk=935 J(100 kg)(9.8 m/s2)(1.5 m)cos30°=0.73$

Therefore, the coefficient of the sliding friction is $0.73_$ .

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Answer 4

Question

Chapter 6, Problem 100P

(a)

To determine

The amount of energy dissipated by friction.

(a)

Expert Solution

Answer to Problem 100P

The amount of energy dissipated by friction is $935 J_$ .

Explanation of Solution

Figure 1 represents the free body diagram of the crate.

COLLEGE PHYICS, Chapter 6, Problem 100P

For an isolated system the total energy of the system (the sum of mechanical energy and its internal energy) is conserved.

$ΔK+ΔU+ΔEint=0$

Here, $ΔK$ is the change in kinetic energy, $ΔU$ is the change in potential energy, and $ΔEint$ is the change in internal energy of the system.

In this situation, the work done by the frictional force is equal to the change in internal energy of the system.

The above equation is reduced to.

$(12mvf2−12mvi2)+mgΔy+ΔEint=0$

Here, $ΔEint$ is the energy dissipated by friction, $m$ is the mass of the crate, $g$ is the acceleration due to gravity, and $Δy$ is the distance travelled along the incline.

Since the final velocity of the crate is zero and the distance travelled along incline is given by $−dsinθ$ , the above equation is reduced to

$−12mvi2−mgdsinθ+ΔEint=0$ (I)

Here, $d$ is the distance travelled along the incline.

Rearrange the above equation.

$ΔEint=12mvi2+mgdsinθ$ (II)

Conclusion:

Substitute $100 kg$ for $m$ , $2 m/s$ for $vi$ , $9.8 m/s2$ for $g$ , $1.5 m$ for $d$ , and $30°$ for $θ$ in equation (I), to find $ΔEint$ .

$ΔEint=12(100 kg)(2 m/s)2−(100 kg)(9.8 m/s2)(1.5 m)sin30°=935 J$

Therefore, the amount of energy dissipated by friction is $935 J_$ .

(b)

To determine

The coefficient of the sliding friction.

(b)

Expert Solution

Answer to Problem 100P

The coefficient of the sliding friction is $0.73_$ .

Explanation of Solution

Write the expression for net force acting along y direction.

$∑Fy=0$

Apply the above condition in Figure 1.

$N−mgcosθ=0N=mgcosθ$ (III)

Write the expression for internal energy dissipated by friction.

$ΔEfric=μkNd$

Here, $μk$ is the coefficient of kinetic friction, $N$ is the normal force, and $d$ is the distance travelled along the incline.

Use the above equation in equation (III).

$ΔEfric=μkmgcosθd$

Rearrange the above equation to find $μk$ .

$μk=ΔEfricmgdcosθ$ (IV)

Conclusion:

Substitute $935 J$ for $ΔEint$ , $100 kg$ for $m$ , $9.8 m/s2$ for $g$ , $1.5 m$ for $d$ , and $30°$ for $θ$ in equation (IV), to find $μk$ .

$μk=935 J(100 kg)(9.8 m/s2)(1.5 m)cos30°=0.73$

Therefore, the coefficient of the sliding friction is $0.73_$ .

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Answer 5

Chapter 6, Problem 100P

(a)

To determine

The amount of energy dissipated by friction.

(a)

Expert Solution

Answer to Problem 100P

The amount of energy dissipated by friction is $935 J_$ .

Explanation of Solution

Figure 1 represents the free body diagram of the crate.

COLLEGE PHYICS, Chapter 6, Problem 100P

For an isolated system the total energy of the system (the sum of mechanical energy and its internal energy) is conserved.

$ΔK+ΔU+ΔEint=0$

Here, $ΔK$ is the change in kinetic energy, $ΔU$ is the change in potential energy, and $ΔEint$ is the change in internal energy of the system.

In this situation, the work done by the frictional force is equal to the change in internal energy of the system.

The above equation is reduced to.

$(12mvf2−12mvi2)+mgΔy+ΔEint=0$

Here, $ΔEint$ is the energy dissipated by friction, $m$ is the mass of the crate, $g$ is the acceleration due to gravity, and $Δy$ is the distance travelled along the incline.

Since the final velocity of the crate is zero and the distance travelled along incline is given by $−dsinθ$ , the above equation is reduced to

$−12mvi2−mgdsinθ+ΔEint=0$ (I)

Here, $d$ is the distance travelled along the incline.

Rearrange the above equation.

$ΔEint=12mvi2+mgdsinθ$ (II)

Conclusion:

Substitute $100 kg$ for $m$ , $2 m/s$ for $vi$ , $9.8 m/s2$ for $g$ , $1.5 m$ for $d$ , and $30°$ for $θ$ in equation (I), to find $ΔEint$ .

$ΔEint=12(100 kg)(2 m/s)2−(100 kg)(9.8 m/s2)(1.5 m)sin30°=935 J$

Therefore, the amount of energy dissipated by friction is $935 J_$ .

(b)

To determine

The coefficient of the sliding friction.

(b)

Expert Solution

Answer to Problem 100P

The coefficient of the sliding friction is $0.73_$ .

Explanation of Solution

Write the expression for net force acting along y direction.

$∑Fy=0$

Apply the above condition in Figure 1.

$N−mgcosθ=0N=mgcosθ$ (III)

Write the expression for internal energy dissipated by friction.

$ΔEfric=μkNd$

Here, $μk$ is the coefficient of kinetic friction, $N$ is the normal force, and $d$ is the distance travelled along the incline.

Use the above equation in equation (III).

$ΔEfric=μkmgcosθd$

Rearrange the above equation to find $μk$ .

$μk=ΔEfricmgdcosθ$ (IV)

Conclusion:

Substitute $935 J$ for $ΔEint$ , $100 kg$ for $m$ , $9.8 m/s2$ for $g$ , $1.5 m$ for $d$ , and $30°$ for $θ$ in equation (IV), to find $μk$ .

$μk=935 J(100 kg)(9.8 m/s2)(1.5 m)cos30°=0.73$

Therefore, the coefficient of the sliding friction is $0.73_$ .

Answer 6

Chapter 6, Problem 100P

(a)

To determine

The amount of energy dissipated by friction.

(a)

Expert Solution

Answer to Problem 100P

The amount of energy dissipated by friction is $935 J_$ .

Explanation of Solution

Figure 1 represents the free body diagram of the crate.

COLLEGE PHYICS, Chapter 6, Problem 100P

For an isolated system the total energy of the system (the sum of mechanical energy and its internal energy) is conserved.

$ΔK+ΔU+ΔEint=0$

Here, $ΔK$ is the change in kinetic energy, $ΔU$ is the change in potential energy, and $ΔEint$ is the change in internal energy of the system.

In this situation, the work done by the frictional force is equal to the change in internal energy of the system.

The above equation is reduced to.

$(12mvf2−12mvi2)+mgΔy+ΔEint=0$

Here, $ΔEint$ is the energy dissipated by friction, $m$ is the mass of the crate, $g$ is the acceleration due to gravity, and $Δy$ is the distance travelled along the incline.

Since the final velocity of the crate is zero and the distance travelled along incline is given by $−dsinθ$ , the above equation is reduced to

$−12mvi2−mgdsinθ+ΔEint=0$ (I)

Here, $d$ is the distance travelled along the incline.

Rearrange the above equation.

$ΔEint=12mvi2+mgdsinθ$ (II)

Conclusion:

Substitute $100 kg$ for $m$ , $2 m/s$ for $vi$ , $9.8 m/s2$ for $g$ , $1.5 m$ for $d$ , and $30°$ for $θ$ in equation (I), to find $ΔEint$ .

$ΔEint=12(100 kg)(2 m/s)2−(100 kg)(9.8 m/s2)(1.5 m)sin30°=935 J$

Therefore, the amount of energy dissipated by friction is $935 J_$ .

Answer 7

Chapter 6, Problem 100P

(a)

To determine

The amount of energy dissipated by friction.

Answer 8

(a)

Expert Solution

Answer to Problem 100P

The amount of energy dissipated by friction is $935 J_$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Figure 1 represents the free body diagram of the crate.

COLLEGE PHYICS, Chapter 6, Problem 100P

For an isolated system the total energy of the system (the sum of mechanical energy and its internal energy) is conserved.

$ΔK+ΔU+ΔEint=0$

Here, $ΔK$ is the change in kinetic energy, $ΔU$ is the change in potential energy, and $ΔEint$ is the change in internal energy of the system.

In this situation, the work done by the frictional force is equal to the change in internal energy of the system.

The above equation is reduced to.

$(12mvf2−12mvi2)+mgΔy+ΔEint=0$

Here, $ΔEint$ is the energy dissipated by friction, $m$ is the mass of the crate, $g$ is the acceleration due to gravity, and $Δy$ is the distance travelled along the incline.

Since the final velocity of the crate is zero and the distance travelled along incline is given by $−dsinθ$ , the above equation is reduced to

$−12mvi2−mgdsinθ+ΔEint=0$ (I)

Here, $d$ is the distance travelled along the incline.

Rearrange the above equation.

$ΔEint=12mvi2+mgdsinθ$ (II)

Conclusion:

Substitute $100 kg$ for $m$ , $2 m/s$ for $vi$ , $9.8 m/s2$ for $g$ , $1.5 m$ for $d$ , and $30°$ for $θ$ in equation (I), to find $ΔEint$ .

$ΔEint=12(100 kg)(2 m/s)2−(100 kg)(9.8 m/s2)(1.5 m)sin30°=935 J$

Therefore, the amount of energy dissipated by friction is $935 J_$ .

Answer 13

(b)

To determine

The coefficient of the sliding friction.

(b)

Expert Solution

Answer to Problem 100P

The coefficient of the sliding friction is $0.73_$ .

Explanation of Solution

Write the expression for net force acting along y direction.

$∑Fy=0$

Apply the above condition in Figure 1.

$N−mgcosθ=0N=mgcosθ$ (III)

Write the expression for internal energy dissipated by friction.

$ΔEfric=μkNd$

Here, $μk$ is the coefficient of kinetic friction, $N$ is the normal force, and $d$ is the distance travelled along the incline.

Use the above equation in equation (III).

$ΔEfric=μkmgcosθd$

Rearrange the above equation to find $μk$ .

$μk=ΔEfricmgdcosθ$ (IV)

Conclusion:

Substitute $935 J$ for $ΔEint$ , $100 kg$ for $m$ , $9.8 m/s2$ for $g$ , $1.5 m$ for $d$ , and $30°$ for $θ$ in equation (IV), to find $μk$ .

$μk=935 J(100 kg)(9.8 m/s2)(1.5 m)cos30°=0.73$

Therefore, the coefficient of the sliding friction is $0.73_$ .

Answer 14

(b)

To determine

The coefficient of the sliding friction.

Answer 15

(b)

Expert Solution

Answer to Problem 100P

The coefficient of the sliding friction is $0.73_$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Write the expression for net force acting along y direction.

$∑Fy=0$

Apply the above condition in Figure 1.

$N−mgcosθ=0N=mgcosθ$ (III)

Write the expression for internal energy dissipated by friction.

$ΔEfric=μkNd$

Here, $μk$ is the coefficient of kinetic friction, $N$ is the normal force, and $d$ is the distance travelled along the incline.

Use the above equation in equation (III).

$ΔEfric=μkmgcosθd$

Rearrange the above equation to find $μk$ .

$μk=ΔEfricmgdcosθ$ (IV)

Conclusion:

Substitute $935 J$ for $ΔEint$ , $100 kg$ for $m$ , $9.8 m/s2$ for $g$ , $1.5 m$ for $d$ , and $30°$ for $θ$ in equation (IV), to find $μk$ .

$μk=935 J(100 kg)(9.8 m/s2)(1.5 m)cos30°=0.73$

Therefore, the coefficient of the sliding friction is $0.73_$ .

Answer 20

Ch. 6.2 - Prob. 6.2CP Ch. 6.2 - Prob. 6.1PP Ch. 6.2 - Find the tension if Diane pulls at an angle θ =...Ch. 6.3 - Prob. 6.3PP Ch. 6.3 - Prob. 6.3ACP Ch. 6.3 - Prob. 6.3BCP Ch. 6.4 - Prob. 6.4ACP Ch. 6.4 - You toss a basketball straight up and then catch...Ch. 6.4 - A ball thrown straight up at an initial speed of...Ch. 6.5 - Prob. 6.6PP

The amount of energy dissipated by friction.

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The amount of energy dissipated by friction.

Answer to Problem 100P

Explanation of Solution

The coefficient of the sliding friction.

Answer to Problem 100P

Explanation of Solution

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