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A constant horizontal force of 28 N is exerted by a string attached to a 8-kg block being pulled across a tabletop. The block also experiences a frictional force of 6 N due to contact with the table.
- a. What is the horizontal acceleration of the block?
- b. If the block starts from rest, what will its velocity be after 3 seconds?
- c. How far will it travel in these 3 seconds?
(a)
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The horizontal acceleration of the block if a horizontal force of 28 N is exerted by a string attached to a block being pulled across a tabletop and it also experiences a frictional force of 6 N due to contact with the table.
Answer to Problem 1SP
The horizontal acceleration of the block if a horizontal force of 28 N is exerted by a string attached to a 8 kg block being pulled across a tabletop is 2.75 m/s2.
Explanation of Solution
Given info: The horizontal force is 28 N, the mass of the block is 8 kg and frictional force is 6 N.
Write the expression for the net horizontal force.
Fnet=Ftension−Ffriction
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 28 N for Ftension and 6 N for Ffriction in the above equation to get Fnet.
Fnet=28 N−6 N=22 N
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 22 N for Fnet and 8 kg for m in the above equation to get a.
a=22 N8 kg=2.75 m/s2
Conclusion:
Thus, the horizontal acceleration of the block if a horizontal force of 28 N is exerted by a string attached to a 8 kg block being pulled across a tabletop is 2.75 m/s2.
(b)
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Velocity of the block after 3 seconds from rest.
Answer to Problem 1SP
Velocity of the block after 3 seconds from rest is 8.25 m/s.
Explanation of Solution
Given info: The time after which velocity is to be find is 3 s.
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 0 m/s2 for v0 ,3 s for t and 2.75 m/s2 for a in the above equation to get v.
v=0 m/s2+(2.75 m/s2)(3 s)=8.25 m/s
Conclusion:
Thus, the velocity of the block after 3 seconds from rest is 8.25 m/s.
(c)
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The distance it travels in 3 seconds.
Answer to Problem 1SP
The distance it travels in 3 seconds is 12.4 m.
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 0 m/s for v0 ,2.75 m/s2 for a and 3 s for t in the above equation to get s.
d=(0 m/s)(3 s)+12(2.75 m/s2)(3 s)2=12.375 m=12.4 m
Conclusion:
Thus, the distance it travels in 3 seconds is 12.4 m.
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Chapter 4 Solutions
Physics of Everyday Phenomena
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