Question

a constant force acts on an object of mass 5 kg for a duration of 5c . it increases the object's velocity from 4 m s-1 to 14 m s -1 . find the magnitude of the applied force. now if the force was applied for a duration of 5c, what would be the final velocity of the object ?​

Answer 1

Answer:

Given,

mass=5kg

t

1

=5s

Initial velocity u=4m/s

Final velocity v=14m/s

t

2

=5s

So,

Let the Force be F

Let the acceleration be a

So,

a=

t

(v−u)

=

2

(14−4)

=10m/s

2

So the magnitude of the applied force is 10N

And the final velocity after 5s is v

So,

v=u+at

v=4+2×7

v=18m/s

The final velocity after 5s is 18m/s

Answer 2

Answer:

force=10 joule

Explanation:

F=ma

F=5(v-u) /t [ a = (change in velocity) /time]

F=5(14-4)/5

F=10j