Question

A constant force acts on an object mass 5kg for a duration of 2s. It increases the objects velocity from 3 m s -^1 to 7 m s-^1 . 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

as change in velocity took place in 2 sec so acceleration is (7-3)/2=2
so force is m.a = 5.2=10

and now the velocity is 7 and if force is applied for 5 seconds it causes 10m/s increase in velocity is final velocity is 17m/s

Answer 2

F = m*(v - u)/t
F = 5*(7 - 3)/2
F = 10 N

a = F/m
a = 10/5
a = 2 m/s^2

v = u + at
v = 7 + 2*5
v = 17 m/s
Final velocity of the object is 17 m/s