Question

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

Answer 1

Given,
m = 5 kg (mass)
t =2 sec (duration for which the Force was applied)
u =3 m/s (initial velocity)
v =7 m/s (final velocity)
To find,
Force (F) by which it act on the object
We know what,
F=ma where a is the acceleration and
a=(v-u)/t
So,
a=(7-3)/2
=4/2
=2m/s²

So,
F=5×2
=10 N
Now,
Force is applied for 5 sec
But,
F=ma
→10=4×(v-3)/5
→10=(4v-12)/5
→10×5=4v-12
→50=4v-12
→50+12=4v
→62=4v
→v=62/4
→v=15.50 m/s

Therefore,
10N of force was applied on the object and when the force was applied for 5 sec their final velocity will be 15.5 m/s

Hope it helps you.Thank you.