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
TPS:
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
Answered by
61
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) = = (-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 + at²
⇒ s = 3×2 - ×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.
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) = = (-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 + at²
⇒ s = 3×2 - ×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.
Answered by
2
Answer:
3m
Explanation:
this is the correct answer
thank you
Attachments:
Similar questions