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# 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 - CBSE (Science) Class 11 - Physics

#### Question

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 = 0, the position of the body at that time to be = 0, and predict its position at t = –5 s, 25 s, 100 s.

#### Solution 1

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 "m/s"^2

(i) At t = -5s

Acceleration, a' = 0 and u = 10 m/s

s = ut + 1/2 a't^2

= 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 xx 25 + 1/2 xx (-20) xx (25)^2

= 250 + 6250 =- 6000 m

(iii) At t = 100 s

For 0 <= t <= 30

a = –20 m/s2

u = 10 m/s

S1 = ut + 1/2 a"t2

= 10 xx 30 + 1/2 xx (-20) xx (30)^2

= 300 - 9000

= -8700 m

For 30' < t <= 100s

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

#### Solution 2

Here m = 0.40 kg, u = 10 ms^(-1), F = -8 N(retarding force)

As a = F/m = - 8/0.4 = - 20 ms^(-2)

Also  S = ut + 1/2 at^2

1) Position at t = - 5s

S = 10(-5) + 1/2 xx 0 xx (-5)^2 = -50 m

2) Position at t = 25 s

S_1 = 10 xx 25 + 1/2 xx (-20)xx(25)^2 = -6000 m  = - 6 km

3) Postion at t = 100 s

S_2 = 10xx30 + 1/2 xx (-20) xx (30)^2 = -8700 m

At t = 30 s, v = u  + at

v = 10 - 20 xx 30 = -590 ms^(-1)

Now for motion from 30 s to 100s

S_3 = -590 xx  70 + 1/2 (0) xx (70)^2 = -41300 m

"Total distance" = S_2 + S_3 = -8700 - 41300 = -50000 m = - 50 km

Is there an error in this question or solution?

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Solution 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 Concept: The Law of Inertia.
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