Question

A 6 kg block slides down a ramp which is at an incline of 15°. If the frictional force is 5.40 N, what is the coefficient of friction? Assume g = 10 m/s2 At what incline (measured in degrees) will the box slide at a constant velocity?

Answer 1

Part A.

Friction Force along the incline will be given by:

Ff = uk*N

N = Normal force = W*cos A = m*g*cos A

Given that Ff = 5.40 N

m = mass of block = 6 kg

A = incline angle = 15 deg

uk = coefficient of friction = ?

Ff = uk*m*g*cos A

uk = Ff/(m*g*cos A)

uk = 5.40/(6*10*cos 15 deg)

uk = 0.093 = Coefficient of friction

Part B.

Now the box will be sliding at constant velocity when it's net acceleration is zero.

From Newton's 2nd law:

Fnet = m*a = m*0 = 0

Now Net force along the incline will be:

Fnet = (m*g*sin A) - Ff

m*g*sin A = uk*m*g*cos A

tan A = uk

A = arctan (uk) = arctan (0.093)

A = 5.31 deg

Please Upvote.