Question

State Newton’s second law of motion. Derive, its mathematical expression. (b) A constant force acts on an object of mass 5 KG for duration of 2 second, it increases the velocity of the 3m/s to 7m/s. Find the force actIng on the object. (1+2+2)​

Answer 1

Answer:

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