Question

A body of mass 0.40 kg moving initially with a constant speed of 10 m s to the north is
subject to a constant force of 8.0 N directed towards the south for 30 s. Take the instant
the force is applied to be t = 0, the position of the body at that time to be x = 0, and predict its
position at t= -100s

Answer 1

Answer:

Given:

The mass of body, m=0.40kg

Constant initial speed of the body is, u=10m/s

Force that acts on the body is, F=8.0N

(i)

At t= –5 s

The force starts acting on the body from t=0 s.

So, the acceleration of the body during this time was zero and it moves at a constant speed.

Position of the body is given by:

x=v×t

x=10×(−5)

=−50 m

(ii)

At t=25 s

Since the force acts in the opposite direction of motion of particle. So, the acceleration of the body due to the force acting on it is:

a=

M

F

=

0.40

−8.0

=−20ms

−2

The position is given by the second equation of motion:

s

′

=ut

′

+

2

1

a"t

2

=10×25+

2

1

×(−20)×25

2

=250−6250=−6000m