Question

11. A constant force acts on an object of mass 2 points 6 kg for a duration of 3 s. It increases the object's velocity from 3 m/s to 9 m/s. The magnitude of the applied force is If the force was applied for a duration of 5 s, the final velocity of the object is » 10 N, 5m/s 10 N, 13 m/s 12 N, 5 m/s O 12 N, 13 m/s​

Answer 1

Given,

mass=5kg

t  

1

​

=2s

Initial velocity u=3m/s

Final velocity v=7m/s

t  

2

​

=5s

So,

Let the Force be F

Let the acceleration be a

So,

a=  

t

(v−u)

​

=  

2

(7−3)

​

=2m/s  

2

 

 

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