Physics
Physics
5th Edition
ISBN: 9781260486919
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 7, Problem 109P
To determine

The speed of the boy at the bottom.

Expert Solution & Answer
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Answer to Problem 109P

The speed of the boy at the bottom is 13m/s_.

Explanation of Solution

Write the expression for the conservation of energy of the boy alone.

Ki+Ui+Wnc=Kf+Uf (I)

Here, Ki is the initial kinetic energy, Ui is the initial potential energy, Wnc is the work done before the second friend jumps to the sled, Kf is the final kinetic energy and Uf is the final potential energy

Write the expression for the initial kinetic energy.

Ki=0

Write the expression for the initial potential energy.

Ui=m1gh1=m1gd1sinθ

Here, m1 is the mass of the boy, g is the acceleration due to gravity and h1 is the initial height of the boy along the slope.

Write the expression to find the work done before the second friend jumps to the sled.

Wnc=μm1gd1cosθ

Write the expression for the final kinetic energy.

Kf=12m1v12

Write the expression for the final potential energy.

Uf=0

Rewrite equation (I) to find the velocity of the boy.

0+m1gd1sinθ+μm1gd1cosθ=12m1v12+0v1=2gd1(sinθμcosθ) (II)

Write the expression for conservation of momentum and rewrite it to find the initial velocity of both the boys.

m1v1=(m1+m2)v2v2=m1v1(m1+m2) (III)

Write the expression for the conservation of energy of the two boys.

Ki+Ui+Wfriction=Kf+Uf (IV)

Write the expression for the initial kinetic energy of two boys.

Ki=12(m1+m2)v22

Write the expression for the initial potential energy of two boys.

Ui=(m1+m2)gh2

Write the expression to find the work done after the second friend jumps to the sled.

Wfriction=μ(m1+m2)gd2cosθ

Write the expression for the final kinetic energy.

Kf=12(m1+m2)v32

Here, v3 is the final velocity.

Write the expression for the final potential energy.

Uf=0

Rewrite equation (IV) to find the final velocity of the boys.

12(m1+m2)v22+(m1+m2)gh2+μ(m1+m2)gd2cosθ}=12(m1+m2)v32+0v3=2gd2(sinθμcosθ)+v22 (V)

Conclusion:

Substitute 9.8m/s2 for g, 20m for d1, 0.12 for μ and 15° for θ in equation (II).

v1=2(9.8m/s2)(20m)(sin15°(0.12)cos15°)=7.48m/s

Substitute 7.48m/s for v1, 60kg for m1 and 50kg for m2 in equation (III).

v2=(60kg)(7.48m/s)(60kg+50kg)=4.08m/s

Substitute 9.8m/s2 for g, 50m for d2, 4.08m/s for v2, 0.12 for μ and 15° for θ in equation (V).

v3=2(9.8m/s2)(50m)(sin15°(0.12)cos15°)+(4.08m/s)2=13m/s

Therefore, the speed of the boy at the bottom is 13m/s_.

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Chapter 7 Solutions

Physics

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