Force acts for 20 s on a body of mass 20 kg, starting from rest,… — Solution | JEE Physics

MCQEasyJEE 2023Motion in a Straight Line

Force acts for $20 \, \text{s}$ on a body of mass $20 \, \text{kg}$ , starting from rest, after which the force ceases, and then the body describes $50 \, \text{m}$ in the next $10 \, \text{s}$ . The value of force will be:

A
$40 \, \text{N}$
B
$5 \, \text{N}$
C
$20 \, \text{N}$
D
$10 \, \text{N}$

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: A body of mass $20 \, \text{kg}$ starts from rest. A force acts for $20 \, \text{s}$ , then ceases. In the next $10 \, \text{s}$ , the body covers $50 \, \text{m}$ .

Find: The value of the force.

After the force ceases, the body moves with constant velocity. Therefore,

S = vt

50 = v \times 10

v = 5 \, \text{m/s}

Using

v = u + at

with $u = 0$ during the first $20 \, \text{s}$ ,

5 = a \times 20

a = \frac{1}{4} \, \text{m/s}^2

Now,

F = ma

F = 20 \times \frac{1}{4} = 5 \, \text{N}

Therefore, the force is $5 \, \text{N}$ and the correct option is B.

Common mistakes

Assuming the body continues to accelerate during the next $10 \, \text{s}$ is incorrect because the force has already ceased. After that instant, the motion is with constant velocity, so use $S = vt$ for that interval.
Using the total time $30 \, \text{s}$ in $v = u + at$ is wrong. The acceleration acts only for the first $20 \, \text{s}$ , so the correct substitution is $5 = a \times 20$ .
Applying $F = ma$ directly with $a = \frac{50}{10}$ confuses velocity with acceleration. First find velocity from the uniform-motion interval, then compute acceleration from the earlier forced motion.

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Answer

Step-by-step solution

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