Question

A constant force acts for 10sec , on a stationary body of mass 10kg after which the force ceases to act. The body moves a distance of 50m in next 5sec. The magnitude of force is

Answer 1

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Magnitude of force = 10N

${\huge{\boxed{\mathcal{\red{Explanation}}}}}$

It is given that body was Stationary thus we can say

Initial velocity of body = u = 0m/s

Force applied = ?

Final velocity = v = ?

Mass of body = 10kg

Acceleration = ?

Time = 5, 10

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Coming to question:-

Since force ceases to act after 10 secs and then body moves 50m in 5 secs with its particular velocity. Thus the final velocity of body will be :-

$v = \frac{d}{t}$

$v = \frac{50}{5} = 10m</p><p>/s$

Now we have final velocity of body is 10m/s.

Now what about acceleration?

Acceleration of a body is represented by :-

$final \: velocity - initial \: velocity</p><p>/t$

$= \frac{v - u}{t}$

After inputting the values in this equation we get :

$a = \frac{10 - 0}{10} \\ \\ a = \frac{10}{10} \\ \\ a = 1m</p><p>/ {s}^{2}$

Now according to newtons second law of motion

$f = mass \: \times acceleration$

Thus force applied in this body will be :

$f = 10 \times 1 \\ \\ f = 10newton$

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BY SCIVIBHANSHU

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