Why does the mass not matter?

Both the driving force (mg sin θ) and the friction force (μk mg cos θ) are proportional to m, so m cancels when you divide by m. A 1 kg and a 50 kg block with the same μk accelerate at the same 3.20 m/s^2.

How do I know the block actually slides?

A block starting from rest slides only if tan θ is greater than the coefficient of static friction. Here tan 30° = 0.577. If the problem says it slides, use μk; if it gives μs and asks whether it moves, compare μs with 0.577.

Physics 1: 5.0 kg Block on a 30° Incline, μk = 0.20: Find the Acceleration (Free-Body Diagram)

The Problem

A 5.0 kg block slides down a ramp inclined at 30° above the horizontal. The coefficient of kinetic friction between the block and the ramp is 0.20. Use $g = 9.8\ \text{m/s}^2$.

(a) Find the acceleration of the block.

(b) If the block starts from rest, how fast is it moving after sliding 2.0 m along the ramp?

Step 1: Draw the Free-Body Diagram With Tilted Axes

Three forces act on the block:

Weight $mg$, straight down
Normal force $N$, perpendicular to the ramp surface
Kinetic friction $f_k$, up the slope (opposite the motion)

Choose the x-axis along the incline (positive = down the slope) and the y-axis perpendicular to it. Only gravity has to be split:

Along the slope: $mg\sin\theta$
Into the slope: $mg\cos\theta$

Step 2: Perpendicular Direction (No Acceleration)

$$ N = mg\cos 30^\circ = (5.0)(9.8)(0.866) = 42.4\ \text{N} $$

Step 3: Friction Force

$$ f_k = \mu_k N = 0.20 \times 42.4 = 8.49\ \text{N} $$

Step 4: Newton's Second Law Along the Slope

$$ mg\sin 30^\circ = (5.0)(9.8)(0.500) = 24.5\ \text{N} $$

$$ F_{\text{net}} = 24.5 - 8.49 = 16.0\ \text{N} $$

$$ a = \frac{F_{\text{net}}}{m} = \frac{16.0}{5.0} = \mathbf{3.20\ m/s^2}\ \text{down the slope} $$

Symbolic version (shows the mass cancels):

$$ a = g(\sin\theta - \mu_k\cos\theta) = 9.8(0.500 - 0.20 \times 0.866) = 9.8(0.327) = 3.20\ \text{m/s}^2 $$

Step 5: (b) Speed After 2.0 m From Rest

Constant acceleration, so use $v^2 = v_0^2 + 2ad$:

$$ v = \sqrt{2(3.20)(2.0)} = \sqrt{12.8} = \mathbf{3.58\ m/s} $$

It takes $t = \sqrt{2d/a} = \sqrt{4.0/3.20} = 1.12$ s to cover that distance.

Numbers at a Glance

Quantity	Value
mg sin 30° (down the slope)	24.5 N
Normal force N = mg cos 30°	42.4 N
Kinetic friction μk N	8.49 N
Net force	16.0 N
Acceleration	3.20 m/s²
Speed after 2.0 m from rest	3.58 m/s

Traps That Cost Points

Setting N = mg. On a ramp, $N = mg\cos\theta$ (42.4 N here, not 49 N). Using 49 N gives friction 9.8 N and a wrong acceleration of 2.94 m/s².
Swapping sine and cosine. The component along the slope is $mg\sin\theta$. Quick check: at θ = 0 (flat floor) the along-slope pull must be zero, and $\sin 0 = 0$.
Friction in the wrong direction. Kinetic friction always opposes the motion. If the block were pushed up the ramp, friction would point down the slope and the deceleration would be $g(\sin\theta + \mu_k\cos\theta)$.
Calculator in radians. $\sin 30$ in radian mode is −0.988.

Variant: Frictionless Ramp

With $\mu_k = 0$: $a = g\sin 30^\circ = 4.90\ \text{m/s}^2$. Friction of 0.20 removes about a third of that.

The same "split the vector into components" step is the core of projectile motion: see the ball kicked at 20 m/s at 30° walkthrough.