Find the coefficient of kinetic friction between a 3.80-kg block and the horizontal surface on which it rests if an 87.0-N/m spring must be stretched by 6.50 cm to pull it with constant speed. Assume that the spring pulls in the horizontal direction.

 μ = 0.15

Explanation:

Let's start by using Hooke's law to find the force applied to the block

          F = k x

          F = 87.0 0.065

          F = 5.655 N

Now we use the translational equilibrium relation since the block has no acceleration

          ∑ F = 0

          F -fr = 0

          F = fr

the expression for the friction force is

          fr = μ N

if we write Newton's second law for the y-axis

          N -W = 0

          N = W = mg

we substitute

          F = μ mg

          μ = F / mg

          μ =

          μ = 0.15

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards