Principles of Physics: A Calculus-Based Text
Principles of Physics: A Calculus-Based Text
5th Edition
ISBN: 9781133104261
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 7, Problem 12P

A crate of mass 10.0 kg is pulled up a rough incline with an initial speed of 1.50 m/s. The pulling force is 100 N parallel to the incline, which makes an angle of 20.0° with the horizontal. The coefficient of kinetic friction is 0.400, and the crate is pulled 5.00 m. (a) How much work is done by the gravitational force on the crate? (b) Determine the increase in internal energy of the crate–incline system owing to friction. (c) How much work is done by the 100-N force on the crate? (d) What is the change in kinetic energy of the crate? (e) What is the speed of the crate after being pulled 5.00 m?

(a)

Expert Solution
Check Mark
To determine

To determine: The work done by the gravitational force on the crate.

Answer to Problem 12P

Answer: The work done by the gravitational force on the crate is 168J .

Explanation of Solution

Explanation:

Given information:

A crate of mass 10kg is pulled up a rough incline with an initial speed of 1.5m/s and the pulling force is 100N parallel to the incline, which makes an angle of 20.0° with the horizontal. The coefficient of kinetic friction is 0.4 , and the crate is pulled 5.00m .

The free body diagram of the system is shown below.

Principles of Physics: A Calculus-Based Text, Chapter 7, Problem 12P

Figure I

The vertical distance moved by the crate is,

h=dsinθ

  • h is the vertical distance.
  • d is the distance moved by crate along the incline.

The formula to calculate the work done by gravitational force is,

Wg=mgh

Substitute dsinθ for h in the above equation.

Wg=mgdsinθ

Substitute 10kg for m , 9.81m/s2 for g , 20° for θ and 5.00m for d to find Wg .

Wg=10kg×(9.81m/s2)×5.00m×sin(20°)168J

Conclusion:

Therefore, the work done by the gravitational force on the crate is 168J .

(b)

Expert Solution
Check Mark
To determine

To determine: The increase in internal energy of the crate-incline system owing to friction.

Answer to Problem 12P

Answer: The increase in internal energy of the crate-incline system owing to friction is 184J .

Explanation of Solution

Explanation:

Given information:

A crate of mass 10kg is pulled up a rough incline with an initial speed of 1.5m/s and the pulling force is 100N parallel to the incline, which makes an angle of 20.0° with the horizontal. The coefficient of kinetic friction is 0.4 , and the crate is pulled 5.00m .

Formula to calculate kinetic frictional force is,

fk=μN

  • μ is the kinetic frictional coefficient.
  • N is the normal force between crate and plane.
  • fk is the kinetic frictional force.

Formula to calculate the increase in the internal energy is,

E=fkd

  • E is the internal energy of the crate.

Substitute μN for fk and mgcosθ for N in above equation.

E=μmgcosθ×d

Substitute 10kg for m , 9.81m/s2 for g , 20° for θ , 0.4 for μ and 5.00m for d to find E .

E=0.4×10kg×9.81m/s2×cos20°×5.00m=184J

Conclusion:

Therefore, the increase in internal energy of the crate-incline system owing to friction is 184J .

(c)

Expert Solution
Check Mark
To determine

To determine: The work done by the 100N force on the crate.

Answer to Problem 12P

Answer: The work done by the 100N force on the crate is 500J .

Explanation of Solution

Explanation:

Given information:

A crate of mass 10kg is pulled up a rough incline with an initial speed of 1.5m/s and the pulling force is 100N parallel to the incline, which makes an angle of 20.0° with the horizontal. The coefficient of kinetic friction is 0.4 , and the crate is pulled 5.00m .

Formula to calculate work done on the crate is,

W=F×d

  • F is the force applied on the crate.

Substitute 100N for F and 5.00m for d to find W .

W=100N×5.00m=500J

Conclusion:

Therefore, the work done by the 100N force on the crate is 500J .

(d)

Expert Solution
Check Mark
To determine

To determine: The change in kinetic energy of the crate.

Answer to Problem 12P

Answer: The change in kinetic energy of the crate is 148J .

Explanation of Solution

Explanation:

Given information:

A crate of mass 10kg is pulled up a rough incline with an initial speed of 1.5m/s and the pulling force is 100N parallel to the incline, which makes an angle of 20.0° with the horizontal. The coefficient of kinetic friction is 0.4 , and the crate is pulled 5.00m .

Formula to calculate change in kinetic energy of the crate is,

ΔKE=Wg+WE

  • ΔKE is the change in kinetic energy of the crate.

Substitute 168J for Wg , 500J for W and 184J for E to find ΔKE .

ΔKE=168J+500J184J=148J

Conclusion:

Therefore, the change in kinetic energy of the crate is 148J .

(e)

Expert Solution
Check Mark
To determine

To determine: The speed of the crate after being pulled 5.00m .

Answer to Problem 12P

Answer: The speed of the crate after being pulled 5.00m is 5.64m/s .

Explanation of Solution

Explanation:

Given information:

A crate of mass 10kg is pulled up a rough incline with an initial speed of 1.5m/s and the pulling force is 100N parallel to the incline, which makes an angle of 20.0° with the horizontal. The coefficient of kinetic friction is 0.4 , and the crate is pulled 5.00m .

Formula to calculate change in kinetic energy of the crate is,

ΔKE=12mvf212mvi2

  • vf is the final speed of the crate.
  • vi is the initial speed of the crate.

Substitute 148J for ΔKE , 10kg for m and 1.5m/s for vi to find vf .

148J=12×10kg×vf212×10kg×(1.5m/s)2vf=2×148J10kg+(1.5m/s)2=5.64m/s

Conclusion:

Therefore, the speed of the crate after being pulled 5.00m is 5.64m/s .

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Chapter 7 Solutions

Principles of Physics: A Calculus-Based Text

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