Question

A force acts for 10 seconds on a body of mass 10 raised to power minus 2 kg initially.ly at rest after which the force ceases to act the body transverse 0.5m in the next 5 second the magnitude of the force is

Answer 1

Let the body starts from rest.
∴u=0
Final velocity = distance / time = 50/5 = 10 m/s
Now, using equation of motion, v = u + at
a=
t
v−u
​
=
10
10−0
​
=1ms
−2

Force = F= ma = 10kg×1ms
−2
= 10 N
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