A horizontal force of 4 N acts on a block of weight of 12 N which is initially at rest on a smooth floor. Find the speed of the block after 6 seconds.
Answers
Answered by
2
Explanation:
The various forces acting on the block is as shown in the figure.
Here, m=2 kg,μ=0.1,F=6 N,g=10 ms
−2
Force of friction,
f=μN=0.1×2 kg×10 ms
−2
=2 N
Net force with which the block moves
F
′
=F−f=6N−2N=4N
Net acceleration with which the block moves
a=
m
F
′
=
2kg
4N
=2 ms
−2
Distance travelled by the block in 10 s is
d=
2
1
at
2
=
2
1
×2 ms
−2
(10)
2
=100 m(∴u=0)
As the applied force and displacement are in the same direction therefore angle between the applied force and the displacement is θ=0
∘
.
Hence, work done by the applied force,
W
F
=Fdcosθ=(6N)(100m)cos0
∘
=600 J.
hope it helps plzz mark as brainliest
Similar questions