Physics
Physics
5th Edition
ISBN: 9781260486919
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 6, Problem 142P

(a)

To determine

The speed of block by considering the energy.

(a)

Expert Solution
Check Mark

Answer to Problem 142P

The speed of block is 2.0m/s.

Explanation of Solution

According to work energy theorem, the work performed by a system will be equivalent to the change in its kinetic energy.

Write an expression for work energy theorem for the block.

Wtotal=ΔKWgrav+Wfriction=12mvf212mvi2mgdsinθμkmgdcosθ=12mvf212m(0)2mgdsinθμkmgdcosθ=12mvf2 (I)

Here, Wtotal is the total work done, ΔK is the change in kinetic energy, Wgrav is the work done by gravity, Wfriction is the work done by the frictional force, m is the mass of the block, g is the acceleration due to gravity, vf is the final velocity of the block, vi is the initial velocity of the block, d is the displacement of the block, μk is the coefficient of kinetic friction, and the inclination is θ.

Rewrite the expression (I) to calculate vf.

vf=2gd(sinθ2μkcosθ) (II)

Conclusion:

Substitute 9.80m/s2 for g, 0.38 for μk, 60.0° for θ and 30.0cm for d in equation (II) to find vf.

vf=2(9.80m/s2)((30.0cm)(1m102cm))(sin(60.0°)2(0.38)cos(60.0°))=2(9.80m/s2)(30.0×102m)(sin(60.0°)2(0.38)cos(60.0°))=2.0m/s

Thus, the speed of block is 2.0m/s.

(b)

To determine

The speed of block using Newton’s second law.

(b)

Expert Solution
Check Mark

Answer to Problem 142P

The speed of block is 2.0m/s.

Explanation of Solution

According to Newton’s second law of motion, the net force acting on a system is the product of mass of the system and the acceleration of the system.

Write an expression for Newton’s second law of motion.

ΣFx=mamgsinθμkmgcosθ=magsinθμkgcosθ=ag(sinθμkcosθ)=a (III)

Here, ΣFx is the net horizontal force on the block and a is the acceleration of the block.

Write an expression to calculate the final velocity of the block.

vf=vi2+2ad=(02)+2(g(sinθμkcosθ))d=2gd(sinθ2μkcosθ) (IV)

Conclusion:

Substitute 9.80m/s2 for g, 0.38 for μk, 60.0° for θ and 30.0cm for d in equation (IV) to find vf.

vf=2(9.80m/s2)((30.0cm)(1m102cm))(sin(60.0°)2(0.38)cos(60.0°))=2(9.80m/s2)(30.0×102m)(sin(60.0°)2(0.38)cos(60.0°))=2.0m/s

Thus, the speed of block is 2.0m/s.

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