Question

A constant force acts on a object of mass 5 kg for a duration of 2s. If the velocity of the object increases from 3ms-7 to 7ms-7 during that period then the magnitude of the
applied force is :​

Answer 1

Answer:

see the attachment

Explanation:

hope the answer will help you

Answer 2

Answer:

Let the force be F,

Mass be M

time for which it is accelerated = 2s

velocity is increased from 3m/s to 7m/s

so, acceleration is , a = (7-3)/2 = 2m/s²

The force applied is, F = ma = (5)(2) = 10 N

Now, the force is applied for 5s, the velocity after 5s will be,

v=u+at

=> v =3+(2)(5)

=>v=13m/s , this is final velocity.

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