Now add a little friction

The question

A $2.0\,\mathrm{kg}$ block is already sliding down a fixed, straight $30^\circ$ slope. The coefficient of kinetic friction is $\mu_k = 0.20$ . Ignore air resistance and take $g = 9.8\,\mathrm{m/s^2}$ .

Which way does kinetic friction act?
Find its magnitude.
Find the acceleration along the slope.

The phrase “already sliding” matters. We are using kinetic friction, not deciding whether static friction can prevent motion from starting.

A hint · Keep the two directions separate

Kinetic friction opposes the block’s motion relative to the surface, so it acts up the slope. Its magnitude is $\mu_k N$ . The perpendicular equation still gives $N = mg\cos\theta$ .

Show the worked solution

The normal force is $N = mg\cos 30^\circ \approx 16.97\,\mathrm{N}$ .

The kinetic friction magnitude is

f_k = \mu_k N \approx 0.20 \times 16.97 \approx 3.39\,\mathrm{N}.

With downhill positive, friction has a negative component:

mg\sin\theta - f_k = ma.

Substituting the expression for friction:

a = g(\sin\theta - \mu_k\cos\theta) \approx 3.2\,\mathrm{m/s^2}.

The acceleration is down the slope, but smaller than the frictionless value of $4.9\,\mathrm{m/s^2}$ .

One more question

If the block were instead sliding up the same slope, would you use the same friction direction?

Check your reasoning

No. Kinetic friction would point down the slope. Both friction and the downhill component of gravity would then act downhill, slowing the upward motion.

While it slides upward, the downhill acceleration would be $g(\sin\theta + \mu_k\cos\theta) \approx 6.6\,\mathrm{m/s^2}$ . When it stops, the kinetic-friction model no longer decides whether it remains at rest; that requires considering static friction.

The question

One more question

Connected learning.