The Physics of Everyday Phenomena
The Physics of Everyday Phenomena
8th Edition
ISBN: 9780073513904
Author: W. Thomas Griffith, Juliet Brosing Professor
Publisher: McGraw-Hill Education
Question
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Chapter 4, Problem 1SP

(a)

To determine

The horizontal acceleration of the block if a horizontal force of 30N is exerted by a string attached to a block being pulled across a tabletop and it also experiences a frictional force of 5N due to contact with the table.

(a)

Expert Solution
Check Mark

Answer to Problem 1SP

The horizontal acceleration of the block if a horizontal force of 30N is exerted by a string attached to a 5kg block being pulled across a tabletop is 5m/s2.

Explanation of Solution

Given info: The horizontal force is 30N, the mass of the block is 5kg and frictional force is 5N.

Write the expression for the net horizontal force.

Fnet=FtensionFfriction

Here,

Fnet  is the net force acting on the block

Ftension is the horizontal force

Ffriction is the frictional force

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

Substitute 30N for Ftension and 5N for Ffriction in the above equation to get Fnet.

Fnet=30N5N=25N

Write the expression for the acceleration of the horizontal acceleration of the block.

a=Fnetm

Here,

m is the mass of the block

a is the horizontal acceleration of the block

Substitute 22N for Fnet and 8kg for m in the above equation to get a.

a=25N5kg=5.0m/s2

Conclusion:

Thus, the horizontal acceleration of the block if a horizontal force of 30N is exerted by a string attached to a 5kg block being pulled across a tabletop is 5.0m/s2.

(b)

To determine

Velocity of the block after 3 seconds from rest.

(b)

Expert Solution
Check Mark

Answer to Problem 1SP

Velocity of the block after 3 seconds from rest is 15m/s.

Explanation of Solution

Given info: The time after which velocity is to be find is 3s.

Write the expression for the equation of motion of the block.

v=v0+at

Here,

v is the velocity of the block after t second

v0 is the initial velocity

a is the acceleration

Substitute 0m/s2 for v0 ,3s for t and 5.0m/s2 for a in the above equation to get v.

v=0m/s2+(5.0m/s2)(3s)=15m/s

Conclusion:

Thus, the velocity of the block after 3 seconds from rest is 15m/s.

(c)

To determine

The distance it travels in 3 seconds.

(c)

Expert Solution
Check Mark

Answer to Problem 1SP

The distance it travels in 3 seconds is 22.5m.

Explanation of Solution

Write the expression for the distance travelled by the block.

d=v0t+12at2

Here,

d is the distance

a is the acceleration

t is the time

Substitute 0m/s for v0 ,5.0m/s2 for a and 3s for t in the above equation to get s.

d=(0m/s)(3s)+12(5.0m/s2)(3s)2=22.5m

Conclusion:

Thus, the distance it travels in 3 seconds is 22.5m.

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Chapter 4 Solutions

The Physics of Everyday Phenomena

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