HOW FAST IS A BLOCK MOVING AT THE BOTTOM OF A SMOOTH SLOPE?
On a smooth slope the speed at the bottom depends only on the vertical height dropped, not on the length of the slope or its angle. Use v = √(2gh) with h the vertical drop, which for a 10 m slope at 30° is 5 m, giving 10 m/s.
A block is released from rest at the top of a smooth slope 10 m long, inclined at 30° to the horizontal. Taking g = 10 m/s2, what is its speed at the bottom?
Use energy, not forces. The only quantity that matters is how far the block has fallen vertically, and the 10 m is measured along the slope, not downwards.
- vertical drop h = 10 sin 30° = 10 × ½ = 5 m
- energy: ½mv2 = mgh, so v2 = 2gh
- v2 = 2 × 10 × 5 = 100
- v = 10 m/s
14 m/s is what you get if you use the slope length as the height putting h = 10 instead of 5. That gives v² = 200 and v ≈ 14.1 m/s. The number in the question is a distance along the slope, and the angle is supplied precisely so that you have to convert it. If the angle were not needed it would not be there.
50 m/s is what you get if you stop at v² and never take the root v² = 100 is not the speed. Under time pressure this is a genuinely common slip, which is why the un-rooted value is usually one of the options.
For anything falling freely or sliding on a smooth surface, v = √(2gh) with h the vertical drop, and nothing else about the path matters. A block down a slope, a ball down a curved ramp and a stone dropped off the edge all arrive at the same speed from the same height. Recognising that lets you skip resolving forces entirely.
THE GENERAL RULE
Choose energy over forces whenever the question gives you a start and an end but does not ask about anything in between. Forces give you acceleration, which you then have to integrate; energy gives you the speed directly. And whenever a distance is quoted along a slope, the first thing to write down is its vertical component, because that is the only part that does work against gravity.
Updated · written to the published ESAT content specification