Question

A constant force acting on an object of 5 kg for a period of 10 second increases the velocity of an object from 3 metre per second to 7 M per second find the magnitude of the applied force now if the force be applied for a period of 5 second what would be the final velocity of the object?


explain in easy manner

Answer 1

Let the Force be F
mass = 5 kg
time(t) = 10 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)/10 = 0.4 m/s²

So the magnitude of the applied force,
F= m*a
​​​​​​ =5*0.4
=2 N

and the final velocity after 5 s is v

so v = u + at
v = 3 + 0.4 x 5
v = 5 m/s

The final velocity after 5 s is 5 m/s