Question

a block of mass 2 kg is initially at rest on a horizontal floor, moves under the action of a force of 10N. The coefficient of friction between the block and the floor is 0.2. If g=10 m/s². find work done by applied force, frictional force and net force and change in kinetic energy in 4s is?​

Answer 1

Answer:

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Explanation:

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Answer 2

Answer:

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= mF ′ = 2kg4N =2 ms −2

Distance travelled by the block in 10 s is

d= 21 at 2 = 21 ×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.