Question

A 5 kg block is sliding down a plane inclined at 30^0 with a constant velocity of 4 m/s. To determine the coefficient of friction between the block and the plane, what pieces of information do you need?

Answer 1

Answer:

The necessary information is if the forces acting on the block are in equilibrium

The coefficient of friction is 0.577

Explanation:

Where the forces acting on the object are in equilibrium, we have;

At constant velocity, the net force acting on the particle = 0

However, the frictional force is then given as

F = mg sinθ

Where:

m = Mass of the block

g = Acceleration due to gravity and

θ = Angle of inclination of the slope

F = 5×9.81×sin 30 = 24.525 N

Therefore, the coefficient of friction is given as

24.525 N = μ×m×g × cos θ = μ × 5 × 9.81 × cos 30 = μ × 42.479

μ × 42.479 N= 24.525 N

∴ μ = 24.525 N ÷ 42.479 N = 0.577

Answer 2

Answer:

We only need the angle of inclination.

The coefficient of friction is 0.58

Explanation:

Check attachment for diagram

For a book placed on an inclined plane, the force acting on the object causing it to slide downwards is the moving Force (Fm).

Fm = Wsin(theta) = mgsin(theta)

The force opposing the moving force is the frictional force Ff

Ff = nR where

n is the coefficient of friction

R is the normal reaction = mgcos(theta)

For the body to have a constant velocity (not accelerate), then Fm =Ff and W = R

n = Ff/R

n = Fm/R

n = Wsintheta/Wcostheta

n = sintheta/costheta

n = tan(theta)

Given theta = 30°

n = tan 30°

n = 0.58

the coefficient of friction between the block and the plane is 0.58

The calculation shows that the only information we need to get the coefficient of friction is the angle is inclination