Question

A force acting on a particle of mass 200g displaces it through 400 cm in 2s. If the initial velocity of the particle is zero, then the magnitude of the force?

Answer with explaination

Answer 1

GIVEN:-

• $\tt{Mass\:of\:Particle = 200g = 0.2kg}$

• $\tt{Displacement = 400cm = 4m}$

• $\tt{Initial\:Velocity = 0m/s}$

• $\tt{Time\:taken = 2s}$

TO FIND:-

• The Magnitude of Force.

FORMULAE USED:-

• ${\boxed{\tt{ F = ma}}}$

• ${\boxed{\tt{ S = ut + \dfrac{1}{2}\times{a}\times{(t)^2}}}}$

Where,

S = Distance

u = initial Velocity

t = Time

m = Mass

a = Acceleration

Now,

$\implies\tt{ S = ut + \dfrac{1}{2}\times{a}\times{(t)^2}}$

$\implies\tt{ 4 = 0 + \dfrac{1}{2}\times{a}\times{(2)^2}}$

$\implies\tt{ 4 = \dfrac{1}{\cancel{2}}\times{a}\times{\cancel{4}}}}$

$\implies\tt{ 4 = 2a}$

$\implies\tt{ a = 2m/s^{-2}}$

Therefore, Using F =ma,

$\implies\tt{ F = ma}$

$\implies\tt{ F = 0.2\times{2}}$

$\implies\tt{ F = 0.4N}$

Hence, The Magnitude of the force applied is 0.4N. It is very small.

Answer 2

Answer

Magnitude of the force = 0.4 N

Given

A force acting on a particle of mass 200 g displaces it through 400 cm in 2 s

To Find

If the initial velocity of the particle is zero, then the magnitude of the force is

Formula Applied

2nd equation of motion

$\to \rm s=ut+\dfrac{1}{2}at^2$

Newton's Second Law

$\to \rm F=ma$

Solution

Mass , m = 200 g = 0.2 Kg

Displacement , s = 400 cm = 4 m

Time , t = 2 s

Initial velocity , u = 0 m/s

Apply 2nd equation of motion ,

$\to \rm s=ut+\dfrac{1}{2}at^2\\\\\to \rm 4=(0)(2)+\dfrac{1}{2}a(2)^2\\\\\to \rm 4=0+2a\\\\\to \rm a=2\ m/s^2$

Apply 2nd law of Newton sir ,

$\to \rm F=ma\\\\\to \rm F=(0.2)(2)\\\\\to \rm F=0.4\ N$

So , Magnitude of the force = 0.4 N