Energy Conservation on a Frictionless Ramp

On a frictionless ramp, the only work done on the ball is by gravity. By conservation of energy, all the gravitational potential energy at the top converts to kinetic energy at the bottom. The shape of the ramp (steep, curved, gentle) doesn't matter — only the height does.

The Calculation

$mgh = \tfrac{1}{2}mv^{2}$

The mass cancels:

$v = \sqrt{2gh}$

This is the same formula as for free fall — and that's the key insight: gravity is a conservative force, so the work it does depends only on the change in height, not the path taken.

Step-by-Step Solution

Given: $h = 5\;\text{m}$, $g = 9.81\;\text{m/s}^{2}$, frictionless, initially at rest.

Find: The speed at the bottom.

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Step 1 — Set up the energy equation.

At the top: $E_1 = mgh + 0 = mgh$ (only PE).

At the bottom: $E_2 = 0 + \tfrac{1}{2}mv^{2}$ (only KE).

By conservation: $mgh = \tfrac{1}{2}mv^{2}$.

Step 2 — Cancel mass and solve.

$gh = \tfrac{1}{2}v^{2}$

$v^{2} = 2gh = 2(9.81)(5) = 98.1\;\text{m}^{2}/\text{s}^{2}$

Step 3 — Take the square root.

$v = \sqrt{98.1} \approx 9.905\;\text{m/s}$

Step 4 — Convert to km/h.

$v = 9.905 \times 3.6 \approx 35.66\;\text{km/h}$

About 22 mph — fast enough to be exhilarating on a real-world ramp.

Step 5 — Compare to a vertical drop.

A ball dropped from $h = 5\;\text{m}$ also reaches $v = \sqrt{2gh} \approx 9.905\;\text{m/s}$. The path doesn't matter — only the height. This is the path-independence of conservative forces.

Step 6 — What if the ramp had friction?

If friction does $W_{\text{fric}}$ joules of negative work, the bottom KE is reduced by that amount:

$\tfrac{1}{2}mv^{2} = mgh - W_{\text{fric}}$

For example, if $\mu_k = 0.2$ on a 30° ramp of length $\ell = h/\sin 30° = 10\;\text{m}$, the friction force is $\mu_k mg\cos\theta = 0.2 \cdot mg\cos 30° \approx 0.173\,mg$. Work done over 10 m is $1.73\,mg$ joules per kg of mass — significant!

The frictionless answer is an upper bound — adding friction always reduces the bottom speed.

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Answer:

$\boxed{\;v = \sqrt{2gh} = \sqrt{2(9.81)(5)} \approx 9.905\;\text{m/s}\;}$

The ball reaches the bottom at about 9.9 m/s (≈ 35.7 km/h or 22 mph), regardless of the ramp's shape or the ball's mass — all that matters is the 5-meter drop in height.

Try It

• Adjust the height — speed scales as $\sqrt{h}$ (not linearly).
• The energy bar on the right shows KE growing as PE shrinks during the descent.
• Try the same height with different ramp shapes (mentally) — the bottom speed is always the same, just the time to reach it differs.