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In a body of mass $m$ at rest for $t$ seconds, a constant force acts on it and then ceases to act. In the next $t$ sec, the body travels a distance of $x$ . Then the magnitude of the force is__

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mx / t^{2}

mx / t

mxt

mx t^{2}

answer is A.

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Detailed Solution

In a body of mass

m

at rest for

t

seconds a constant force acts on it and then ceases to act. In next

t

sec the body travels a distance

x

, magnitude of force is

mx / t^{2} .

Given:

Mass = m

Time = t sec

Distance = x

Initial velocity,

u = 0

Let the velocity of the body after

t sec = v

We know if the initial velocity of a body is

u

and it is under uniform acceleration. Then its velocity

v

at any time given by

v = u + at

------ (1)
Also, using Newton’s first law, we have

F = ma

Therefore

a = \frac{F}{m}

----- (2)
Put (2) in (1)

v = 0 + \frac{F}{m} t

\Rightarrow v = \frac{Ft}{m}

Given the distance covered in another

t seconds

distance = Speed \times time

\Rightarrow x = vt

∴ x = \frac{Ft}{m} t

\Rightarrow F = \frac{mx}{t^{2}}

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