Physics
Physics
5th Edition
ISBN: 9781260486919
Author: GIAMBATTISTA
Publisher: MCG
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Question
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Chapter 6, Problem 36P

(a)

To determine

The speed of the brick when it reaches edge of the roof.

(a)

Expert Solution
Check Mark

Answer to Problem 36P

The speed of the brick when it reaches edge of the roof is 4.43 m/s .

Explanation of Solution

Write the equation for the change in height.

Δy=lsinθ (I)

Here, Δy is the change in height, l is the distance to the edge of the roof, and θ is the angle of inclination of the roof

Write the principle of conservation of energy.

Ki+Ui=Kf+Uf

Here, Ki is the initial kinetic energy of the brick, Kf is the final kinetic energy of the brick, Ui is the initial potential energy of the brick and Uf is the final potential energy of the brick

Since the brick starts from rest, its initial kinetic energy is zero and also its potential energy at the edge of the roof is zero.

Substitute 0 for Ki and Uf in the above equation.

0+Ui=Kf+0Ui=Kf (II)

Write the equation for Ui .

Ui=mgΔy (III)

Here, m is the mass of the brick and g is the acceleration due to gravity

Write the expression for Kf .

Kf=12mv2 (IV)

Here, v is the constant speed of the brick

Put equations (III) and (IV) in equation (II).

mgΔy=12mv2v2=2gΔyv=2gΔy (V)

Conclusion:

Given that the distance to the edge of the roof is 2.00 m and the angle of inclination of the roof

 is 30.0° . The value of acceleration due to gravity is 9.80 m/s2 .

Substitute 2.00 m for l and 30.0° for θ in equation (I) to find Δy .

Δy=(2.00 m)sin30.0°=1.00 m

Substitute 9.80 m/s2 for g and 1.00 m for Δy in equation (V) to find v .

v=2(9.80 m/s2)(1.00 m)=4.43 m/s

Therefore, the speed of the brick when it reaches edge of the roof is 4.43 m/s .

(b)

To determine

The speed of the brick when it reaches edge of the roof if the coefficient of kinetic friction is 0.10 .

(b)

Expert Solution
Check Mark

Answer to Problem 36P

The speed of the brick when it reaches edge of the roof with coefficient of kinetic friction 0.10 is 4.03 m/s .

Explanation of Solution

Write the equation for the work done by the friction force on the brick.

Wf=Flcos180°=Fl (VI)

Here, Wf is the work done by the friction force on the brick and f is the magnitude of force of friction

Write the equation for F .

F=μmgcosθ

Here, μ is the coefficient of kinetic friction

Put the above equation in equation (VI).

Wf=μmglcosθ

This work is stored as the potential energy.

Write the equation for Ui .

Ui=mgΔyμmglcosθ

Put equation (I) in the above equation.

Ui=mglsinθμmglcosθ=mgl(sinθμcosθ) (VII)

Put equations (IV) and (VII) in equation (II).

12mv2=mgl(sinθμcosθ)v2=2gl(sinθμcosθ)v=2gl(sinθμcosθ) (VIII)

Conclusion:

Substitute 9.80 m/s2 for g , 2.00 m for l , 30.0° for θ and 0.10 for μ in equation (VIII) to find v .

v=2(9.80 m/s2)(2.00 m)(sin30.0°0.10cos30.0°)=4.03 m/s

Therefore, the speed of the brick when it reaches edge of the roof with coefficient of kinetic friction 0.10 is 4.03 m/s .

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Chapter 6 Solutions

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