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"How long does the block take to reach the bottom?" looks like a one-line question, but students lose marks on it in three places: assuming the block slides at all, getting the friction term wrong, and forgetting that a block thrown up a rough slope takes different times going up and coming down. This post works one incline from every angle.
Where is this in the syllabus?
The NTA Syllabus for JEE (Main) 2026 lists "static and kinetic friction, laws of friction" in Unit 3, and the NEET (UG) 2026 syllabus has the same line. The NCERT Class 11 chapter Laws of Motion (Reprint 2026-27) shows that a block on an incline starts to slide when tan θ exceeds μs, and that this angle "depends only on μs and is independent of the mass of the block". The time to slide combines that with the equations of motion.
How do you find the acceleration?
Take axes along and perpendicular to the slope, as OpenStax University Physics, section 6.2 does for a snowboarder. For a block of mass m on a slope at angle θ:
Perpendicular: N = mg cos θ
Along the slope, once it is sliding: ma = mg sin θ − μk N
So a = g(sin θ − μk cos θ). Mass cancels. From rest, L = ½at², so t = √(2L/a).
Worked example with sample numbers
A block is released from rest at the top of a 4.0 m long incline at 30°. Take μs = 0.25, μk = 0.20 and g = 9.8 m/s².
Does it slide? tan 30° = 0.577, which is more than μs = 0.25, so yes.
Acceleration: a = 9.8(0.500 − 0.20 × 0.866) = 9.8 × 0.327 = 3.20 m/s².
Time: t = √(2 × 4.0 / 3.20) = √2.50 = 1.58 s.
Speed at the bottom: v = at = 3.20 × 1.58 = 5.06 m/s. Check: v² = 2aL = 25.6, so v = 5.06 m/s.
On a smooth incline the same block would have a = 9.8 × 0.5 = 4.9 m/s² and take √(8.0/4.9) = 1.28 s. Friction stretches the time by a factor of 1.58/1.28 = 1.24.
How do you find μ from the time ratio?
A favourite exam version: "A block takes n times as long to slide down a rough incline as down a smooth one of the same angle. Find μk." Since t is proportional to 1/√a, the accelerations are in the ratio n²:
sin θ / (sin θ − μk cos θ) = n², which gives μk = tan θ (1 − 1/n²).
Check it on our example: n = 1.237, so μk = 0.577 × (1 − 1/1.530) = 0.577 × 0.346 = 0.200. For the textbook case of 45° and n = 2, μk = 1 × (1 − 1/4) = 0.75.
Why is the trip up quicker than the trip down?
Now push the same block up the incline at 5.0 m/s. Friction acts down the slope on the way up, adding to gravity:
Going up: a = g(sin θ + μk cos θ) = 9.8 × 0.673 = 6.60 m/s², against the motion
Distance before it stops: 5.0² / (2 × 6.60) = 1.89 m
Time up: 5.0 / 6.60 = 0.76 s
It slides back because tan θ > μs. Coming down: a = 3.20 m/s², so time down = √(2 × 1.89 / 3.20) = 1.09 s
Speed on return: 3.20 × 1.09 = 3.48 m/s, less than 5.0 m/s because friction did work both ways
The ratio t_down/t_up = √(6.60/3.20) = 1.44.
Which mistakes should you watch for?
N = mg. On a slope N = mg cos θ.
μs for a sliding block. Use μk once it moves.
Friction in the wrong direction. Friction opposes the motion, so it points down the slope on the way up.
Calculator in radians. sin 30 in radians is −0.99, a common source of nonsense answers.
Not checking for sliding. See the callout above. Our post on static friction and stacked blocks covers the static case in detail.
A practice routine
Derive a = g(sin θ − μk cos θ) from a free-body diagram. Our guide to free-body diagrams shows how to draw one.
Redo the example at 37° with μk = 0.25. (Check: a ≈ 3.94 m/s², t ≈ 1.42 s.)
Derive μk = tan θ(1 − 1/n²) yourself, then test it at 60° with n = 2.
Solve one up-and-back question and find both times.
If you keep getting the set-up wrong rather than the algebra, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning, so you can compare your free-body diagram with a full solution.
Today, solve the 37° version on paper before you look at the check values.
