COLLEGE PHYSICS-CONNECT ACCESS
COLLEGE PHYSICS-CONNECT ACCESS
5th Edition
ISBN: 9781260486834
Author: GIAMBATTISTA
Publisher: MCG
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Question
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Chapter 2, Problem 122P

(a)

To determine

The minimum force required to slide both the blocks together along the floor.

(a)

Expert Solution
Check Mark

Answer to Problem 122P

The minimum force required to slide both the blocks together along the floor is 15.1N_.

Explanation of Solution

Write the expression for normal force acting on system of two blocks.

    N=(m1+m2)g        (I)

Here, N is the normal force, m1 is the mass of block 1, and m2 is the mass of block 2, and g is the acceleration due to gravity.

The minimum force required to slide both the blocks together along the floor is equal to the maximum static frictional force.

Write the expression for maximum static frictional force.

    fstat=μsN        (II)

Here, μs is the coefficient of static friction, and fstat is the maximum static frictional force.

Use equation (I) in equation (II).

    fstat=μs(m1+m2)g        (III)

Conclusion:

Substitute 5kg for m1, 2kg for m2, 0.22 for μs, and 9.81m/s2 for g in equation (III), to find fstat.

    fstat=(0.22)(5kg+2kg)(9.81m/s2)=15.1N

Therefore, the minimum force required to slide both the blocks together along the floor is 15.1N_.

(b)

To determine

The maximum force required to push the top block that starts to slide off the lower block.

(b)

Expert Solution
Check Mark

Answer to Problem 122P

The maximum force required to push the top block that starts to slide off the lower block is 33.35N_.

Explanation of Solution

Write the expression for frictional force acting between the blocks.

    f=μsm1g        (IV)

Here, f is the frictional force acting between the blocks, μs is the static frictional force, m1 is the mass of the upper block, and g is the acceleration due to gravity.

Write the expression for net force acting on the system.

    F=μk(m1+m2)g        (V)

Here, F is the net force acting on the system, μk is the coefficient of kinetic friction.

Write the expression for maximum force required to push the top block that starts to slide off the lower block.

    Fmax=F+f        (VI)

Here, Fmax is the maximum force required to push the top block that starts to slide off the lower block.

Use equation (IV) and (V) in equation (VI).

    Fmax=μk(m1+m2)g+μsm1g        (VII)

Conclusion:

Substitute 0.200 for μk, 5kg for m1, 2kg for m2, 9.81m/s2 for g, 0.4 for μs in equation (VII), to find Fmax.

    Fmax=(0.200)(5kg+2kg)(9.81m/s2)+(0.4)(5kg)(9.81m/s2)=33.35N

Therefore, the maximum force required to push the top block that starts to slide off the lower block is 33.35N_.

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Chapter 2 Solutions

COLLEGE PHYSICS-CONNECT ACCESS

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