Physics of Everyday Phenomena
Physics of Everyday Phenomena
9th Edition
ISBN: 9781259894008
Author: W. Thomas Griffith, Juliet Brosing Professor
Publisher: McGraw-Hill Education
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Chapter 4, Problem 2SP

A rope exerts a constant horizontal force of 350 N to pull a 40-kg crate across the floor. The velocity of the crate is observed to increase from 1 m/s to 9 m/s in a time of 2 seconds under the influence of this force and the frictional force exerted by the floor on the crate.

  1. a. What is the acceleration of the crate?
  2. b. What is the net force acting upon the crate?
  3. c. What is the magnitude of the frictional force acting on the crate?
  4. d. What force would have to be applied to the crate by the rope in order for the crate to move with constant velocity? Explain.

(a)

Expert Solution
Check Mark
To determine

The acceleration of the crate.

Answer to Problem 2SP

The acceleration of the crate is 4m/s2 in the direction of velocity.

Explanation of Solution

Given info: The horizontal force is 350N, initial velocity is 1m/s ,final velocity is 9m/s and time for the change in velocity is 2s.

Write the equation of motion of the crate.

v=v0+at

Here,

v is the velocity of the crate

v0 is the initial velocity of the crate

a is the acceleration of the crate

t is the time

Rearrange the above equation to get a.

a=vv0t

Substitute 1m/s for v0 ,9m/s for v and 2s for t in the above equation to get a.

a=9m/s1m/s2s=4m/s2

The direction of the crate is in direction of velocity.

Conclusion:

Thus, the acceleration of the crate is 4m/s2 in the direction of velocity

(b)

Expert Solution
Check Mark
To determine

The net force acting on the crate.

Answer to Problem 2SP

The net force acting on the crate 160N in the direction of the velocity.

Explanation of Solution

Given info: The mass of the crate is 40kg.

Write the expression for the net force.

Fnet=ma

Here,

Fnet is the net force on the crate

m is the mass of the crate

a is the net acceleration of the crate

Substitute 40kg for m and 4m/s2 for a in the above equation to get Fnet.

Fnet=(40kg)(4m/s2)=160N

According to the second law motion this force will be in the direction of acceleration. This implies that the force is in the direction of velocity.

Conclusion:

Thus, the velocity of the block after 3 seconds from rest is 8.25m/s.

(c)

Expert Solution
Check Mark
To determine

The magnitude of the frictional force acting on the crate.

Answer to Problem 2SP

The magnitude of the frictional force acting on the crate is 190N.

Explanation of Solution

Write the expression for the net horizontal force.

Fnet=FconstantFfriction

Here,

Fnet  is the net force acting on the block

Fconstant is the horizontal force

Ffriction is the frictional force

The negative sign indicate that frictional force is opposite to horizontal force

Substitute 350N for Fconstant and 160N for Fnet in the above equation to get Ffriction.

160N=350NFfrictionFfriction=350N160N=190N

Conclusion:

Thus, the magnitude of the frictional force acting on the crate is 190N.

(d)

Expert Solution
Check Mark
To determine

The force that must apply to crate by the rope in order for the crate to move with constant velocity.

Answer to Problem 2SP

The force that must apply to crate by the rope in order for the crate to move with constant velocity is 190N

Explanation of Solution

Write the expression for the net horizontal force.

Fnet=FconstantFfriction

The frictional force is 190N. The crate will move with a constant velocity if the acceleration is zero.

Substitute 0N for Fnet and 190N for Ffriction in the above equation to get Fconstant.

0N=Fconstant190NFconstant=190N

Conclusion:

Thus, the force that must apply to crate by the rope in order for the crate to move with constant velocity is 190N

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Chapter 4 Solutions

Physics of Everyday Phenomena

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