Block sliding down a 30° incline with friction: find the acceleration
The question: A block slides down a 30° slope with friction coefficient μ = 0.2. Find its acceleration (g = 9.8 m/s²).
The block accelerates down the slope at about 3.20 m/s², from a = g(sin θ − μ cos θ).
A block slides down a 30° incline with μ = 0.2. See the force diagram, Newton's second law along the slope, why N = mg cos θ, and the answer a ≈ 3.20 m/s², explained step by step.
Step-by-step transcript
The question
A block slides down a thirty degree slope with friction. Let's find its acceleration, step by step.
The situation
Picture a block on a rough wooden ramp.
What we know and the plan
We know the slope is thirty degrees, the friction coefficient is zero point two, and g is nine point eight. We need the acceleration down the slope. We'll use Newton's second law along the slope, because only those forces change the motion.
The forces
Three forces act on the block. Gravity pulls straight down, the normal force pushes out of the slope, and friction pushes back up the slope.
Forces along the slope
The net force down the slope is the part of gravity along the slope, m g sine theta, minus friction, which pushes back up. Friction is mu times the normal force, and on a slope the normal force is m g cos theta. Newton's second law says m a equals the net force, with friction swapped in. Every term has an m, so the mass cancels. a equals g times sine theta minus mu cos theta.
Put in the numbers
Now the numbers. g is nine point eight, theta is thirty degrees and mu is zero point two. Sine thirty is a half, and cos thirty is about zero point eight six six. Inside the bracket, that's about zero point three two seven. Multiply, and the acceleration is about three point two metres per second squared.
The answer
So the block speeds up at about three point two metres per second squared.
Checking the answer
Quick check. With no friction it would be g sine theta, four point nine. Friction should slow it down, and three point two is less. That makes sense.
The common mistake
Watch out: on a slope the normal force is not m g. It's m g cos theta.
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