Question

A body of mass 500 g, initially at rest, is acted upon
by a force which causes it to move a distance of 4
m in 2 s, Calculate the force applied.
​

Answer 1

GIVEN:-

• $\textrm{Mass\:of\:body = 500g = 0.5kg}$

• $\rm{Initial\:Velocity =0m/s}$

• $\rm{Distance\: Covered = 4m}$

• $\rm{ Time\:taken = 2s}$.

TO FIND:-

• The force applied.

FORMULAE USED:-

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

• ${\huge{\boxed{\rm{F = ma}}}}$.

Where,

S = Distance

u = initial Velocity

a = Acceleration

t = time

F = force

m = mass

Now,

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

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

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

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

$\implies\rm{ a= \dfrac{4}{2}}$

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

Therefore,

$\implies\rm{ F = ma}$

$\implies\rm{ F = 0.5\times{2}}$

${\implies{\huge{\boxed{\underline{\rm{ F = 1N}}}}}}$.

Hence, The force Applied is 1N.

Answer 2

$\hugel\underline{\overline{\mid{\bold{\red{Given-}}\mid}}}$

▪Mass- 500g= 5kg

▪Initial velocity- 0m/s

▪Distance- 4m

▪Time- 2s

》Find-》

▪Force

$\hugel\underline{\overline{\mid{\bold{\orange{Solution-}}\mid}}}$

Formula of force is

●F=ma●

firstly we will find the acceleration [a]

By using 2nd law of motion

$\boxed{s = ut + \frac{1}{2} {a}t^{2} }$

putting values in this we get,

$4 = 0 + \frac{1}{2} a \times 2 \times 2$

$a = 2 \frac{m}{{s}^{2}}$

 

putting the value of a in the F=ma formula,

F=ma

$f = 0.5 \times 2 = 1$

   

$\boxed{F=1N}$