Question

A constant force acts on a body of mass 5kgfor a duration of 2s. It increases the object's velocity from 3m/s to 7m/s . Find the magnitude of the force that was applied for a duration of 5s, what would be the final velocity of the object?

Answer 1

Let the Force be F
mass = 5 kg
time(t) = 2 s
initial velocity(u) = 3 m/s
final velocity(v) = 7 m/s
so let the acceleration be a
so a = (v - u)/t = (7 - 3)/2 = 2 m/s²

F=ma
F=5×2=10 N

So the magnitude of the applied force is 10 N

and the final velocity after 5 s is v

so v = u + at
v = 3 + 2 × 5
v = 13 m/s

The final velocity after 5 s is 13 m/s