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. | Learn NCERT solution | Education portal Class 11 Physics | Study online Unit-5 Laws Of Motion



Q.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.

 

Answer:-

Mass of the body, m = 0.40 kg

Initial speed of the body, u = 10 m/s due north
Force acting on the body, F = –8.0 N
Acceleration produced in the body, a = F / m = -8.0 / 0.40 = -20 ms-2

(i) At t = –5 s
Acceleration, a‘ = 0 and u = 10 m/s
s = ut + (1/2) a’ t2
= 10 × (–5) = –50 m



(ii) At t = 25 s
Acceleration, a” = –20 m/s2 and u = 10 m/s
s’ = ut’ + (1/2) a” t2
= 10 × 25 + (1/2) × (-20) × (25)2
= 250 – 6250 = -6000 m

(iii) At t = 100 s
For 0 ≤ t ≤ 30 s
a = -20 ms-2
u = 10 m/s
s1 = ut + (1/2)a”t2
= 10 × 30 + (1/2) × (-20) × (30)2
= 300 – 9000  =  -8700 m
For 30 < t ≤ 100 s

As per the first equation of motion, for t = 30 s, final velocity is given as:
v = u + at
= 10 + (–20) × 30 = –590 m/s
Velocity of the body after 30 s = –590 m/s
For motion between 30 s to 100 s, i.e., in 70 s:
s2 = vt + (1/2) a” t2
= -590 × 70 = -41300 m
∴ Total distance, s” = s1 + s2 = -8700 -41300 = -50000 m = -50 km.