Whe coefficient of kinetic frictio.A 300-kg crate is placed on an adjustable inclined plane. As one end of the incline is raised, the crate...

We assume the crate is inclined at an angle ,x to the horizontal. Let us assume the mass of the crate m kg.

Now the force acting on the crate  are: (i) weight force of the crate (ii), the force along the plane of the crate due to its acceleration, a. (iii) th friction  opposing the force of accelatiing movement.

Resolving the weight force along and vertical to the surface of the plane we get:

Force along the plane = mg sinx, where g is the acceleration due to gravitation. But this is equal to ma.

Force normal to the plane = mg cosx

If  M is the coefficient of friction, then , frictional force = M times normal force = M*mg cosx.

given are m = 300kg, a = 0.7 m/s^2,  From scentific tables, g = 9.81m/s^2 .

So, the resultant of the force mg sinx and mg cosx is ma.Or

ma = mg sinx - M*mgcosx Or

a = g(sinx-Mcosx). Or

M cosx = (sinx - a/g). Or

M =(sinx -a/g)/cosx = (sinx - 0.7/9.981)/cosx.

M = (sinx - 0.71355759)/cosx is the coeefficient of friction between the crate and the plane.

So the value of M could be found using the different values of the adjustable inclination angle x with the help of the above equation