Question

a block of mass 2 kg initially at rest moves under the action of an applied horizontal force of 6 Newton on a rough horizontal surface the coefficient of friction between block and surface is 0.1 the work done by the applied force in 10 second is

plss.. ans both the que .​

Answer 1

Answer:

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=10ms−2

Force of friction, f=μN

=0.1×2kg×10ms−2

=2N

Net force with which the block moves

F'=F-f

=6N - 2 N

=4N

Net acceleration with which the block moves

a=F'm

=4N2kg

=2ms−2

Distance travelled by the block in 10 s is

d=12at2

=12×2ms−2(10s)2

=100m (∴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,

WF=Fdcosθ

=(6N)(100m)cos0∘

=600J