COLLEGE PHYSICS-CONNECT ACCESS
COLLEGE PHYSICS-CONNECT ACCESS
5th Edition
ISBN: 9781260486834
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 6, Problem 142P

(a)

To determine

The speed of the block along the incline.

(a)

Expert Solution
Check Mark

Answer to Problem 142P

The speed of the block along the incline is 2m/s_.

Explanation of Solution

Figure 1 represents the motion of the block along the incline.

COLLEGE PHYSICS-CONNECT ACCESS, Chapter 6, Problem 142P , additional homework tip  1

Consider the block is at a height h initially with the angle of inclination θ.

Apply conservation of energy to Figure 1.

    mgh=mgh2+12mv2+fkl

Here, m is the mass of the block, g is the acceleration due to gravity, fk is the frictional force, l is the distance moved along the incline, h is the initial height of the block, v is the speed of the block and h2 is the height of the block at the intermediate stage of motion.

Rearrange the above equation.

    mg(hh2)=12mv2+fkl        (I)

From Figure 1 hh2=lsinθ and the frictional force fk=μkmgcosθ.

Apply the above condition in equation (I). and rearrange to find v.

    mglsinθ=12mv2+μkmglcosθmglsinθμkmglcosθ=12mv2v=2gl(sinθμkcosθ)        (II)

Conclusion:

Substitute 9.8m/s2 for g, 30cm for l, 0.38 for μk, and 60° for θ in equation (II), to find v.

    v=2(9.8m/s2)(30cm×1m102cm)(sin60°μkcos60°)=1.993m/s2m/s

Therefore, the speed of the block along the incline is 2m/s_.

(b)

To determine

The speed of the block along the incline using newton’s second law of motion, to check which method is easier to calculate the speed.

(b)

Expert Solution
Check Mark

Answer to Problem 142P

The speed of the block along the incline using newton’s second law of motion is 2m/s_ , calculating the speed using newton’s second law of motion is much easier.

Explanation of Solution

Figure 2 represents the free body diagram of the block.

COLLEGE PHYSICS-CONNECT ACCESS, Chapter 6, Problem 142P , additional homework tip  2

Write the net force acting on the block in the vertical direction.

    Nmgcosθ=0N=mgcosθ        (III)

Here, N is the normal force.

Write the net force acting on the block in the horizontal direction.

    F=mgsinθμkN        (IV)

Here μk is the coefficient of kinetic friction.

Use equation (III) in (IV).

    F=mgsinθμkmgcosθ        (V)

Write the expression for acceleration of the block.

    a=Fm

Here, a is the acceleration of the block, F is the force experienced by the block along the incline, and m is the mass of the block.

Use equation (V) in the above equation.

    a=mgsinθμkmgcosθm=gsinθμkgcosθ        (VI)

Write the kinematic equation.

    v2u2=2as

Here, v is the final speed of the block, u is the initial speed of the speed, and s is the distance travelled along the incline.

Since the initial speed of the block is zero, the above equation is reduced to

    v2=2asv=2as        (VII)

Conclusion:

Substitute 9.8m/s2 for g, 0.38 for μk, and 60° for θ in equation (VI), to find a.

    a=(9.8m/s2)sin60°(0.38)(9.8m/s2)cos60°=6.625m/s2

Substitute 6.625m/s2 for a, and 30cm for s in equation (VII), to find v.

    v=2(6.625m/s2)(30cm×1m102cm)=1.993m/s2m/s

Therefore, the speed of the block along the incline using newton’s second law of motion is 2m/s_ , calculating the speed using newton’s second law of motion is much easier.

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Chapter 6 Solutions

COLLEGE PHYSICS-CONNECT ACCESS

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