Question

When a constant force acts on an object with a mass of 8 kg for a duration of 3 s. it increases the object's velocity from 4 m/s to 6 m/s . What is the magnitude of the applied force?
A) 5.33 N B) 4.33 N C) 6.33 N D) 3.33 N

Answer 1

Answer:

by the same method you will get your answer

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