COLLEGE PHYSICS-CONNECT ACCESS
COLLEGE PHYSICS-CONNECT ACCESS
5th Edition
ISBN: 9781260486834
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 6, Problem 16P
To determine

The work done by gravity, work done by the normal force, the work done by the friction, and the magnitude of frictional force.

Expert Solution & Answer
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Answer to Problem 16P

The work done by the normal force is zero, the work done by gravity is 6087.42J_, work done by the frictional force is 2337.42J_, and the magnitude of frictional force is 73.04N_.

Explanation of Solution

The work done by the normal force on the skier is zero since the normal force is perpendicular to the direction of motion of the skier.

Write the expression for work done by the gravity.

    Wgrav=mgΔy        (I)

Here, Wgrav is the work done by the gravity, m is the mass, g is the acceleration due to gravity, Δy is the vertical distance.

From Figure 1 the vertical distance is given by

    Δy=0(32.0m)sinθ=(32.0m)sinθ

Use the above equation in equation (I).

    Wgrav=mg(32.0m)sinθ        (II)

The total work done on the skier is equal to sum of work done by the gravity, and work done by the friction.

    WTotal=Wgrav+Wfriction 

Here, WTotal is the total work done on the skier, Wgrav is the work done by the gravity, and Wfriction  is the work done by the friction.

The total work done by the skier is equal to the change in kinetic energy of the skier. The above equation is written as

    12mv2=Wgrav+Wfriction 

Use equation (I) in the above equation and rearrange.

    12mv2=mgΔy+Wfriction Wfriction =12mv2+mgΔy

Since Δy is equal to (32.0m)sinθ the above equation is reduced to

    Wfriction =12mv2mg(32.0m)sinθ        (III)

Write the expression for work done by frictional force.

    Wfric=fkΔx        (IV)

Here, Wfric is the work done by frictional force, fk is the frictional force, and Δx is the distance travelled along the incline.

Rearrange the above equation to find fk.

    fk=WfricΔx        (V)

Conclusion:

Substitute 75.0kg for m, 9.8m/s2 for g, and 15° for θ in equation (II), to find Wgrav.

    Wgrav=(75.0kg)(9.8m/s2)(32.0m)sin15°=6087.42J

Substitute 75.0kg for m, 9.8m/s2 for g, 15° for θ and 10m/s2 for v in equation (III), to find Wfriction.

    Wfriction =12(75.0kg)(10m/s2)2(75.0kg)(9.8m/s2)(32.0m)sin15°=2337.42J

Substitute 2337.42J for Wfriction , and 32.0m for Δx in equation (V), to find fk.

    fk=2337.42J32.0m=73.04N

Therefore, the work done by the normal force is zero, the work done by gravity is 6087.42J_, work done by the frictional force is 2337.42J_, and the magnitude of frictional force is 73.04N_.

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Chapter 6 Solutions

COLLEGE PHYSICS-CONNECT ACCESS

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