Question

A constant force acts on an object of mass 5 Kg for a duration of 2 seconds. It increases the objects v=u+atvelocity from 3m/s to 7m/s . Find the magnitude of the applied force. Now if the tax were applied for a duration of 5seconds, what would be the final velocity of the object ?

Answer 1

The initial and final velocity of the items are given, however we must assume that initial and final velocity. We must calculate the object’s force using the mass and acceleration of the item.

 

 Given that 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

 

So,

a = (v−u)/t

= (7−3)/2 = 2m/s²

 

So the magnitude of the applied force is 10N

And the final velocity after 5s is v 

 

So,

v = u + at

v = 3 + 2 × 5

v = 13m/s

 

The final velocity after 5s is 13m/s.

 

 

Final velocity of the object after 5s is 13 m/s