Question

xample 9.1 A constant force acts on an
object of mass 5 kg for a duration of
2 s. It increases the object's velocity
from 3 m s-1 to 7 m s-1. Find the
magnitude of the applied force. Now, if
the force was applied for a duration of
5 s, what would be the final velocity of
the object?​

Answer 1

Answer:

Step-by-step explanation:

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