Question

A constant force acts on an
object of mass 5 kg for a duration of
2 s. It increases the object's velocity
from 3 m s- to 7 m sl. Find the
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

1st part : F = m × a = m × (v-u)/t = 5 × (7-3)/2 = 20/2 = 10N

2nd part : Keeping the mass and force constant,

F = m × (v-u)/t

=> F × t = m(v-u)

=> 10×5 = 5v - 5×3

=> 50+15 = 5v

=> v = 65/5 = 13 m/s.