Question

d. If the duration for which the force acts is
shortened to 2 seconds what will be the final
velocity of the object.
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Answer 1

Answer:

Explanation:

Given,

mass=5kg

t  1  =2s

Initial velocity u=3m/s

Final velocity v=7m/s

t  2  =5s

So,

Let the Force be F

Let the acceleration be a

So,

a=  t (v−u)  =  2 (7−3)

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=2m/s  2

 

 

So the magnitude of the applied force is 10N

And the final velocity after 5s is v  

So,

v=u+at

v=3+2×5

v=13m/s

The final velocity after 5s is 13m/s