Question

A body of mass 0.25 kg moving with a velocity of 12ms-1 is brought to rest by applying a force of 0.6 N.
Calculate (i) time taken to stop the body and (ii) displacement before comming to rest

Answer 1

GIVEN :  

mass = 0.25 kg

velocity = 12 m/sec

force = 0.6 N

TO FIND -

• time taken to stop the body

• displacement

SOLUTION -

As the body is coming to rest,

        final velocity (v) = 0

  we know, Force(F) = mass(m) * acceleration(a)

                   acceleration (a) = $\frac{F}{m}$

                           a = $\frac{0.6}{0.25}$

                           a = 2.4 m/$sec^{2}$

(i) We know,

v = u + at      ( where, a = acceleration; u = initial velocity; t = time taken)

or, 0 = u - at  ( as retardation is happening)

or, u = at

or, 12 = 2.4 * t

∴   t    =  12/2.4 = 5 sec

(ii) we know,

     $v^{2} = u^{2} + 2as$

   or,  o =    144 - 2 *2.4*s

   or,  4.8s = 144

   or,       s  = $\frac{144}{4.8}$

   or,       s  =  30 m

Therefore, Displacement before coming to rest is 30m.