Question

A force acts for 10 seconds on a body of mass 10^-2 kg, initially at rest, after which the force ceases to act. The body traverses 0.5 m in the next 5 seconds. The magnitude of the force is :
(a) 9.8 N (b) 98 N (c) 10^-2 N (d) 10^-4 N

Aligarh Muslim University +2 Science/Diploma Entrance Test - 2018-19

Answer 1

let the acceleration be a

so, using second equation of motion we have

S= ut +1/2 at^2

here ,
t=10sec
u=0
and s=0.5m


so,

0.5 = 1/2 × a × 100
so,
a= 1/100 m/sec^2


now F=ma

so , F = 10^-2 ×10^-2


so , F will be 10^-4 N

option c is correct

Answer 2

Hey friend, Harish here.

Here is your answer.

Given that,

1) Mass of the object = 10⁻² Kg

2) The time for which the force acts = 10 sec.

3) Distance moved in next 5 sec = 0.5m

To find,

The magnitude of force.

Solution:

Velocity after it starts moving:

⇒ $Velocity\ (v)= \frac{Displacement }{Time}$

⇒ $\frac{0.5}{5} = 0.1 \ m/s$

We know that, 

⇒ $Acceleration\ (a) = \frac{(v-u)}{t}$

Here, u = initial velocity = 0 m/s. & T = 10 sec. V = 0.1 m/s

⇒ $\frac{(0.1-0)}{10} = 0.01 \ m/s^{2}$

Also,

⇒ $Force\ (f)= Mass\times acceleration.$

⇒ $F= 10^{-2} \times 0.01$

⇒ $10^{-2}\times 10^{-2}$

⇒ $10^{-4}\ N$

Therefore the magnitude of force is 10⁻⁴ N
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Hope my answer is helpful to you.