Question

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

Answer 1

Answer:

Given,

mass=5kg

t1 =2s

Initial velocity u=3m/s

Final velocity v=7m/s

t2=5s

So,

Let the Force be F

Let the acceleration be a

a= (v-u)/t = (7-3)/3 = 4/2 = 2m^2

So the magnitude of the applied force is 10N

So

v=u+at

v=3+2×5

v = 13 m/s

So the velocity will be 13 m