Question

A box of 4kg is placed on wooden plank of length of 1.5m which is lying on the ground the plank is lifted from one end along its length so thst it is become inclined it is noted that when the vertical height of the end of the plank from the ground becomes 0.5 m"the box began to slide" find the cofficent friction between the box and plank

Answer 1

Answer:

Explanation:

Given A box of 4 kg is placed on wooden plank of length of 1.5m which is lying on the ground the plank is lifted from one end along its length so thst it is become inclined it is noted that when the vertical height of the end of the plank from the ground becomes 0.5 m  

Given m = 4 kg, distance  s = 1.5 m, h = 0.5 m

Now when the box just begins to slide

     m g cos theta = R, mg sin theta = μ R

So mg sin theta = μ mg cos theta

Now μ = tan theta

           = h / √ s^2 – h^2

           = 0.5 / √ 1.5^2 – 0.5^2

          = 0.5 / √ 2

          = 0.3536

So coefficient of friction will be 0.3536

Answer 2

Answer:

Explanation:

Given m = 4 kg, distance s = 1.5 m, h = 0.5 m

Now when the box just begins to slide

m g cos theta = R, mg sin theta = μ R

So mg sin theta = μ mg cos theta

Now μ = tan theta

=Opp\hyp

=0.5/√2

=0.353553