Question

Example 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-. 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

m=5kg

v=7m/s

u=3m/s

t=2s

F=m (v-u)/t

F=5 (7-3)/2

F=10N

If t=5s

v=(F×t/m)+u

v=(10×5/5)+3

v=13m/s.

Answer 2

Answer:

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