Question

A mass M slides downward along a rough plane surface inclined at angle \Theta\:Θ= 32.51 in degrees relative to the horizontal. Initially the mass has a speed V_0\:V 0 = 6.03 m/s, before it slides a distance L = 1.0 m down the incline. During this sliding, the magnitude of the power associated with the work done by friction is equal to the magnitude of the power associated with the work done by the gravitational force. What is the coefficient of kinetic friction between the mass and the incline?

Answer 1

Answer: μ = 0.8885

Explanation: Force due to friction is calculated as: $F_{f} = \mu.N$

At an inclined plane, normal force (N) is: N = mgcosθ, in which θ=32.51.

Power associated with work done by friction is $P=F_{f}.x$. The variable x is displacement the object "spent its energy".

Power associated with work done by gravitational force is P = mghcosθ, where h is height.

The decline forms with horizontal plane a triangle as draw in the picture.

To determine force due to friction:

$F_{f}.x=mghcos(\theta)$

$F_{f}=\frac{mghcos(\theta)}{x}$

Replacing force:

$\frac{m.g.h.cos(\theta)}{x} = \mu.m.g.cos(\theta)$

$\mu=\frac{h}{x}$

Calculating h using trigonometric relations:

$sin(32.51) = \frac{h}{1}$

h = sin(32.51)

Coefficient of Kinetic friction is

$\mu=\frac{sin(32.51)}{1}$

μ = 0.8885

For these conditions, coefficient of kinetic friction is μ = 0.8885.