Question

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.

Answer 1

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.

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