Question

Question 5.10 A body of mass 0.40 kg moving initially with a constant speed of 10 m s–1 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 = –5 s, 25 s, 100 s.

Chapter Laws Of Motion Page 110

Answer 1

A=force/mass
Therefore,a=-20m/s square
Then apply the equation of motion,
S=ut+1/2 a×tsquare and find S or distance

Answer 2

Velocity of body = 10j { North direction }
a constant force = 8N directed towards the South .
We know ,
F = ma
8 = 0.4 × a
a = 8/0.4 = 20 m/s²
Hence, acceleration = -20j { South direction }
Now ,
S = ut + 1/at²
S = (10j)t + 1/2(-20j)t²
S =(10t)i -10t² j

Position of body at t = -5
From t = -5 to t = 0 the motion of body is unaccelerated motion .(towards North )
Distance covered by body = speed of body × time
= 10 × (-5) = -50 m

Position of the body at t = 25s
Body is accelerated between 0 to 30sec . so,
S = (10t)j - (10t²)j
= 10×25j - (10×625)j
= -6000 j
= 6000m in South direction

Position of body at t = 100s
Body accelerated in 0 to 30sec after that body moves with constant speed .
So,
Position of particle at t = 30sec
S1 = 10× 30j - 10×30²j
= -8700j m
Velocity at t = 30
V = U + at
V = 10j + (-20j)×30
= -590j m/s
Displacement between 30 to 100sec .
S2 = V×t
= -590j × 70
= -41300j m

Hence, position at t = 100sec
S = S1 + S2
= -8700j - 41300j
= -50000j
Hence, position at t = 100sec of body is 50km South direction .