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

1
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.

Answers

2014-10-04T11:34:15+05:30

This Is a Certified Answer

×
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
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.