A block on an incline

The question

A $2.0\,\mathrm{kg}$ block is released on a fixed, straight, frictionless slope at $30^\circ$ above the horizontal. Ignore air resistance and take $g = 9.8\,\mathrm{m/s^2}$ .

Find the block’s acceleration down the slope.
Find the normal force.
Predict what would change if the block had twice the mass.

Draw a force diagram before writing an equation.

Hint 1 · Choose useful axes

Choose one axis down the slope and another perpendicular to it. The block does not accelerate into or away from the fixed straight surface.

Hint 2 · Resolve the weight

The downhill component of weight is $mg\sin 30^\circ$ . The component into the surface is $mg\cos 30^\circ$ . Gravity is still one force.

Show the worked solution

The only forces are weight and the normal force. With downhill positive:

mg\sin\theta = ma \quad\Rightarrow\quad a = g\sin\theta.

At $30^\circ$ :

a = 9.8 \times 0.5 = 4.9\,\mathrm{m/s^2}.

Perpendicular to the slope, the acceleration is zero:

N - mg\cos\theta = 0.

So $N = 2.0 \times 9.8 \times \cos 30^\circ \approx 17\,\mathrm{N}$ .

Doubling the mass leaves the acceleration unchanged, because mass cancels from the first equation. It doubles the normal force.

Check the result

The acceleration is smaller than $g$ , which makes sense: only a component of gravity acts along the slope. The normal force is also smaller than the weight, $19.6\,\mathrm{N}$ .

If your answer was $N = mg$ , revisit the perpendicular direction. When you’re ready, try the same slope with friction.

The question

Check the result

Connected learning.