Force

I think I did this the long way but here goes.

To find the force, we know

$\overrightarrow { F } =\frac { d\overrightarrow { p } }{ dt } =\frac { d }{ dt } (m\overrightarrow { v } )\approx m\frac { \Delta v }{ \Delta t }$

from there we find the force to be 20 N.

The acceleration is just a = F/m, and we know it to be opposing the motion of the mass as the velocity decreases in the time period.

Plug the acceleration (remember the negative sign), the initial velocity (5.0 m/s), and keeping in mind the initial displacement is zero (because we can choose where our coordinates begin) into:

$s={ s }_{ o }+{ v }_{ o }t+\frac { 1 }{ 2 } a{ t }^{ 2 }$

and we get 30 m - 9 m = 21 m.

Hey I don't understand the use to find the force here............

Acceleration =change in velocity/time. So acceleration= -0.5m/s^2. Now V^2 =u^2+2as. on putting values(remember acceleration is negative) we get s=21m

a=dv/dt=-(5-2)/6=-0.5m/s^2 by second equation of motion,i.e. s=ut+0.5at^2 we get s=30-9=21 hence, displacement is 21m

Actually the question is bit confusing, it does not states clearly whether one has to find force or distance.However when i saw unit meters then i understood we have to find distance here.Please correct the question.

Here we have to find acceleration first then we can use v^2-u^2=2as

Its asking the distance,just didn't see the meter

whats the reason of asking force????????