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Friction and inclined planes questions
Static friction adjusts to oppose impending motion up to its limiting value, F ≤ μR, where R is the normal reaction. At limiting equilibrium, and while sliding, F = μR. Add a slope, resolve the weight along and perpendicular to it, and the same model covers both statics and sliding.
6 original questions · 23 marks · the friction and inclined planes notes · Mechanics
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State the friction law F ≤ μR, explaining when the inequality is an equality.
Worked answer
Friction matches whatever force is trying to slide the surfaces past each other, up to a ceiling of μR. Equality holds only at that ceiling, which happens in limiting equilibrium, on the point of slipping, or while the object is actually sliding. Below the ceiling friction is exactly as large as it needs to be and no larger. B1 for the inequality, B1 for the conditions giving equality.A 6 kg box rests on a rough horizontal floor, μ = 0.4. A horizontal force of 20 N is applied. Show that the box does not move, and state the friction force acting.
Worked answer
Vertically, R = mg = 58.8 N, so friction can supply up to μR = 0.4 × 58.8 = 23.52 N. The applied 20 N is below that ceiling, so the box stays put and friction takes the value 20 N, exactly enough to balance. Quoting 23.52 N as the acting friction is the error the question is set to catch. A 'show that' demands the comparison in writing, so state that 20 < 23.52 before concluding. M1 for R = mg, A1 for μR = 23.52 N, M1 for the comparison, A1 for the friction being 20 N.A 6 kg box rests on a rough horizontal floor, μ = 0.4, and the horizontal force applied to it is increased to 30 N. Find the acceleration.
Worked answer
30 N beats the ceiling of 23.52 N, so the box slides and friction locks at its maximum value. The resultant is 30 − 23.52 = 6.48 N, so a = 6.48/6 = 1.08 m s⁻². Once the box is moving, F = μR exactly, and the problem reduces to ordinary Newton's second law. M1 for the resultant, A1 for 6.48 N, A1 for the acceleration.A block is released on a smooth slope inclined at 30°. Find its acceleration, and explain why the mass does not matter.
Worked answer
Resolving along the slope, the only force component is mg sin 30°, so ma = mg sin 30° and a = g sin 30° = 4.9 m s⁻². The mass multiplies both the driving force and the inertia, so it cancels. Every block slides a smooth 30° slope in the same way, whatever it weighs. M1 for resolving along the slope, A1 for the acceleration, B1 for the mass cancelling.A 3 kg block sits on a rough slope at 20°, μ = 0.5. Show that it remains at rest.
Worked answer
Down the slope the pull is mg sin 20° = 3 × 9.8 × sin 20° = 10.1 N. Perpendicular to the slope, R = mg cos 20° = 27.6 N, so friction can supply up to μR = 0.5 × 27.6 = 13.8 N. The ceiling of 13.8 N beats the pull of 10.1 N, so equilibrium is possible and the block stays at rest with friction actually providing 10.1 N up the slope. On a slope R is mg cos θ and never mg. That substitution is the whole point of the question, and writing R = 29.4 N throws away every mark after it. M1 for mg sin 20°, A1 for the friction ceiling of 13.8 N, M1 for the comparison, A1 for the conclusion.A block of mass 2 kg is projected up a rough slope inclined at 25° with an initial speed of 6 m s⁻¹. The coefficient of friction between block and slope is 0.3. Find the distance the block travels up the slope, and determine whether it then slides back down.
Worked answer
Going up, gravity and friction both act down the slope, so ma = mg sin 25° + μmg cos 25° and a = 9.8(sin 25° + 0.3 cos 25°) = 6.81 m s⁻² of deceleration. The mass cancels. Then 0 = 62 − 2 × 6.81 × s gives s = 2.64 m. At the top the block is at rest, so ask whether the down-slope pull can beat the friction ceiling: mg sin 25° = 8.28 N against μmg cos 25° = 5.33 N. It can, so the block slides back. The same test written as an angle is tan 25° = 0.466 > 0.3 = μ, which is the standard criterion for slipping on a rough slope. Coming down, friction reverses and acts up the slope, so the return acceleration is 9.8(sin 25° − 0.3 cos 25°) = 1.48 m s⁻², much gentler than the climb. M1 A1 for the deceleration up the slope, M1 A1 for the distance, M1 A1 for the comparison that settles the slide-back, B1 for the return acceleration.
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