A block of mass 4 kg is kept over a rough horizontal surface.the coefficient of friction between the block and surface is 0.1 at t=0(3i^)m/s velocity is imparted to the block and simultaneously (-2i^)Nforce starts acting on it. its displacement in first 5 seconds is

If (-2i^) forceacts on the body, the friction will also be acting, it will stop in 2 seconds. Check the information if it is correct.



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Mass of block = 4kg
coefficient of friction (μ) = 0.1
at t=0s,
velocity(u) = 3m/s i^
force acting(F) = (-2N) i^
friction (f) = μmg (-i^) = 0.1×4×10 = (-4N) i^ 
Net force = F+f = (-6N) i^

Acceleration (a) =  \frac{(-6N) i^}{4} = (-1.5m/s²) i^

final velocity(v) = 0
time of travel = t
t = (v-u)/a = (0-3)/(-1.5) = 2s
Distance travelled in first 2 seconds(s) = ut +  \frac{1}{2} at²

⇒ s = 3×2 -  \frac{1}{2} ×1.5×2²

⇒ s = 6 - 3 = 3m

At 2s, the body would have travelled 3m. And still (-2N) i^ will be acting but the friction will now act opposite to this and will be equal to (2N) i^. So there would not be any motion of the body.

Thus at the end of 5s, distance travelled will be 3m.

The question is a bit tricky. If you have still any doubt, leave a comment.

1 5 1
why is the final velocity 0?
I needed to check how much time it is taking to stop, is it greater than 5 or less than 5...
if t>5, then i just need to put t=5 in s=ut+(1/2)at^2
And notice that final friction is different than initial friction both in magnitude and direction.