Question

A constant force acts on an object of mass 10 kg for a duration of 4 s. It increases the object’s velocity from 6 m s –1 to 14 m s -1 . Find the magnitude of the applied force. Now, if the force was applied for duration of 10 s, what would be the final velocity of the object?

Answer 1

Answer:

F=20 N

v=34 m/s

Explanation:

a=(v-u)/t

=(14-6)/4

=8/4

=2 m/s^2

F=ma

=10 kg× 2m/s^2

=20 N

Now acceleration will be 2 m/s^2 if same fore is applied for 10 s.

a=(v-u)/t

2=(v-14)/10

20=v-14

v=20+14

v=34 m/s