Physics
Physics
5th Edition
ISBN: 9781260486919
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 4, Problem 121P

(a)

To determine

The acceleration of two blocks after they released from rest.

(a)

Expert Solution
Check Mark

Answer to Problem 121P

Acceleration of block 1 is 3.9m/s2 in right direction.

Acceleration of block 2 is 3.9m/s2 downwards.

Explanation of Solution

Mass of block m1 is 3.0kg and the mass of block m2 is 2.0kg.

The free body diagram is shown below.

Physics, Chapter 4, Problem 121P

Write the equation for net force on block 1 in vertical direction.

Nm1g=0 (I)

Here, the normal reaction force on block 1 is N, mass of block 1 is m1, and the gravitational acceleration is g.

Write the equation for net force on block 1 in horizontal direction.

Tm1a1x=0 (II)

Here, the tension on string is T, acceleration of block in horizontal direction is a1x.

Write the equation for net force on block 2 in vertical direction.

Tm2g=m2a2y (III)

Here, the mass of block 2 is m2 and the acceleration of m2 in vertical direction is a2y.

Write the condition to avoid the shrinking of cords.

a1x=a2y

Here, the acceleration in right direction is taken as positive and in downward direction is taken as negative quantity.

Introduce a new common variable in place of a1x and a2y.

a=a1x

Here, the new variable to denote the acceleration is a.

a=a2y

Rewrite equations

Rewrite equations (II) and (III) in terms of T by substituting a for a1x and a2y.

T=m1aT=m2a+m2g

Equate the right hand sides of above two equations.

m1a=m2a+m2g(m1m2)a=m2ga=m2g(m1m2)

Conclusion:

Substitute 3.0kg for m1, 2.0kg for m2, and 9.8m/s2 for g in the above equation to find a.

a=(2.0kg)(9.8m/s2)(3.0kg2.0kg)=19.6kgm/s21kg=3.9m/s2

The direction of acceleration of each block is same as that of the tension on string connecting the pulley and each block. Tension on string connecting m1 and pulley is in right direction. Thus, acceleration of m1 should be in right direction. Tension on string connecting m2 and pulley is acting vertically downwards. Thus, acceleration of m2 should be vertically downwards.

Therefore, the acceleration of block 1 is 3.9m/s2 in right direction and the acceleration of block 2 is 3.9m/s2 downwards.

(b)

To determine

The velocity of m1 after 1.2s on the assumption that m1 do not run out of room from the table and m2 do not reaches the floor.

(b)

Expert Solution
Check Mark

Answer to Problem 121P

The velocity is 4.7m/s in right direction.

Explanation of Solution

Mass of block m1 is 3.0kg and the mass of block m2 is 2.0kg.

Write the Newton’s equation to find the velocity of m1.

v=u+aΔt

Here, the velocity of m1 is v, initial velocity is u, and the time taken is Δt.

Conclusion:

Substitute 0m/s for u, 3.9m/s2 for a, and 1.2s for Δt in the above equation to find v.

v=0m/s+(3.9m/s2)(1.2s)=4.7m/s

Due to tension on string, m1 moves in right direction. Thus, velocity is acting rightwards.

Therefore, the velocity is 4.7m/s in right direction.

(c)

To determine

The displacement of m1 in 1.2s.

(c)

Expert Solution
Check Mark

Answer to Problem 121P

Displacement is 2.8m rightwards.

Explanation of Solution

Mass of block m1 is 3.0kg and the mass of block m2 is 2.0kg.

Write the Newton’s equation to find the displacement of m1.

d=uΔt+12a(Δt)2 (IV)

Here, the displacement of m1 is d, initial velocity is u, and the time taken is Δt.

Conclusion:

Substitute 0m/s for u, 3.9m/s2 for a, and 1.2s for Δt in the above equation to find d.

d=(0m/s)(1.2s)+12(3.9m/s2)(1.2s)2=2.8m

m1 is moving rightwards.

Therefore, the displacement is 2.8m rightwards.

(d)

To determine

The displacement of m1 and m2 at 0.40s after it is released from their initial positions.

(d)

Expert Solution
Check Mark

Answer to Problem 121P

The displacement of m1 is 0.31 rightwards.

The displacement of m2 is 0.31 downwards.

Explanation of Solution

Mass of block m1 is 3.0kg and the mass of block m2 is 2.0kg.

Conclusion:

Substitute 0m for u, 0.40s for Δt, and 3.9m/s2 for a in equation (IV) to find d for m1.

d=(0m)(0.40s)+12(3.9m/s2)(0.40s)2=0.31m

Substitute 0m for u, 0.40s for Δt, and 3.9m/s2 for a in equation (IV) to find d for m1.

d=(0m)(0.40s)+12(3.9m/s2)(0.40s)2=0.31m

m1 is moving rightwards and m2 moves vertically downwards.

Therefore, the displacement of m1 is 0.31 rightwards and the displacement of m2 is 0.31 downwards.

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Chapter 4 Solutions

Physics

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Drawing Free-Body Diagrams With Examples; Author: The Physics Classroom;https://www.youtube.com/watch?v=3rZR7FSSidc;License: Standard Youtube License