COLLEGE PHYSICS-CONNECT ACCESS
COLLEGE PHYSICS-CONNECT ACCESS
5th Edition
ISBN: 9781260486834
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 6, Problem 100P

(a)

To determine

The amount of energy dissipated by friction.

(a)

Expert Solution
Check Mark

Answer to Problem 100P

The amount of energy dissipated by friction is 935J_.

Explanation of Solution

Figure 1 represents the free body diagram of the crate.

COLLEGE PHYSICS-CONNECT ACCESS, 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.

    (12mvf212mvi2)+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

    12mvi2mgdsinθ+ΔEint=0        (I)

Here, d is the distance travelled along the incline.

Rearrange the above equation.

    ΔEint=12mvi2+mgdsinθ        (II)

Conclusion:

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

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

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

(b)

To determine

The coefficient of the sliding friction.

(b)

Expert Solution
Check Mark

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.

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

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

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

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Chapter 6 Solutions

COLLEGE PHYSICS-CONNECT ACCESS

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