Question

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

Answer 1

Answer:

Explanation:

Let the force be F,

Mass is 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.