Question

A force of 10 N is applied in opposite direction
on a body of mass 5 kg moving at a velocity of
6 ms -1. In how many seconds does it come to
rest:​

Answer 1

Given:

Force applied on body, F= -10 N

(Negative sign shows that force is acting in opposite direction)

Mass of body, m= 5 Kg

Initial velocity of body, u= 6 m/s

Final velocity of body, v= 0 m/s

(Since, it stops after applying force)  

To Find:

Time taken by body to stop

Solution:

We know that,

• Expression of force F is given by

$\pink{\boxed{\bf{F=Mass\times Acceleration}}}$

• According to first equation of motion for constant acceleration,

$\purple{\boxed{\bf{v=u+at}}}$

where,

v is final velocity

u is initial velocity

a is acceleration

t is time taken

$\rule{190}{1}$

Let the acceleration of given body be 'a'

So, force acting on given body is given by

$\longrightarrow\rm{F=m\times a}$

$\longrightarrow\rm{-10=5\times a}$

$\longrightarrow\rm{5a=-10}$

$\longrightarrow\rm{a=\dfrac{-10}{5}}$

$\longrightarrow\rm{a=-2\:m/s^{2}}$

$\rule{190}{1}$

Now, on applying first equation of motion on given body, we get

$\longrightarrow\rm{v=u+at}$

$\longrightarrow\rm{0=6+(-2)t}$

$\longrightarrow\rm{-6=-2t}$

$\longrightarrow\rm{-2t=-6}$

$\longrightarrow\rm{t=\dfrac{-6}{-2}}$

$\longrightarrow\rm\green{t=3\:s}$

Hence, in 3 seconds the body will comes to at rest.