Question

A constant force acts on a object Of mass 5 kg for a duration of 2 s. It increases the object's velocity from 3 m s-¹ to 7 m s-¹. Find the magnitude of the applied force . Now , if 5 s , what would be the final velocity of the object

Answer 1

Answer:

The final velocity after 5s is 13m/s

Explanation:

Let , Force be F

Mass=5kg

time(t) = 2 s

Initial velocity u=3m/s

Final velocity v=7m/s

So,

Let the acceleration be a

So,

a = (v - u)/t

    = (7 - 3)/2

    = 2 m/s²

So the magnitude of the applied force is 10 N

Then, the final velocity after 5s is v

Then, v = u + at  

v = 3 + 2 x 5

v = 13 m/s  

The final velocity after 5 s is 13 m/s.

Hope it helps!!