Question

A 25 kg block set on an inclined plane slides down at constant sleep when the angle of inclination of the lane is 40 degrees.

a) draw the diagram of the block on the inclined plane showing the forces acting on the block

b) calculate the magnitude of the weight and its component vectors

c) What is the net force parallel to the surface acting on the block?

d) What is the magnitude of the force of kinetic (sliding) friction?

e) Determine the coefficient of sliding friction of the block-plane surface

Please help me with d and e.

Answer 1

a) See attached

b) The block's weight has magnitude

mg = (25 kg) (9.8 m/s²) = 245 N

and we split this vector into components acting parallel and perpendicular to the incline, with magnitudes

• parallel : mg sin(40°) ≈ 157 N

• perpendicular : mg cos(40°) ≈ 188 N

Take "down the plane" to be the positive parallel direction, and the direction of the normal force to be the positive perpendicular direction. Then these components of the weight are positive and negative,

• parallel : mg sin(40°) ≈ 157 N

• perpendicular : mg cos(40°) ≈ -188 N

c) The net force acting parallel to the incline is

∑ F[para] = mg sin(40°) - f = 0

where f is the magnitude of kinetic friction. The net force is zero since the block slides at a constant speed.

d) Solve the equation in (c) for f :

f = mg sin(40°) ≈ 157 N

e) The magnitude of kinetic friction f is proportional to the magnitude of the normal force n by a factor of µ, the coefficient of friction.

f = µn

The net force on the block acting perpendicular to the incline is

∑ F[perp] = n - mg cos(40°) = 0

and it follows that

n = mg cos(40°) ≈ 188 N

Solve for µ :

157 N = µ (188 N) ⇒ µ ≈ 0.84