If $µk=0.4$ determine the distance travelled by the block & the time it will take as it comes to rest.

Let be the distance travelled by block before coming to rest

So, Work done by gravitational force $= mgs \sin15 $

Work done by frictional force $= - mgs \cos15 * (0.4) $

Now, as particle is coming to rest, its final KE will be 0

Applying work-energy principle

$0 - 0.5 mu^2 =$ Work done by gravitational force + Work done by frictional force

$$0 - 0.5 (50g) 20^2 = mgs \sin15 - mgs \cos15 * (0.4) $$

Solving, we get $S= 159.9 m $

Now, applying $V^2 = u^2 + 2as \\ 0 = 20^2 + 2a (159.9) \\ a= -1.251 m/s^2 \\ Using v = u+at \\ 0 = 20 - 1.251(t) \\ t = 15.987 \sec \approx 16 sec $

Distance travelled by block is $159.9 \approx 160 m$ & time taken is 16 sec