Final Velocity of a Rough Block on a Fixed Wedge Calculator

Final Velocity of a Rough Block on a Fixed Wedge Calculator

When a rough block slides down a fixed wedge, some energy is lost to friction. The energy balance is given by:
\[ mgh = \frac{1}{2}mv^2 + \mu\,mg\,h\cot\theta \] which leads to:
\[ v = \sqrt{2gh\left(1 – \mu\,\cot\theta\right)} \]

* Enter the vertical drop \( h \) (m), wedge angle \( \theta \) (°), gravitational acceleration \( g \) (m/s²), and friction coefficient \( \mu \). (For sliding, ensure μ < tanθ.)

Step 1: Enter Parameters

Example: 5 m

Example: 30°

Example: 9.81 m/s²

Example: 0.2

Derived Formula: \( v = \sqrt{2gh\left(1 – \mu\,\cot\theta\right)} \)


Practical Example:
For a vertical drop of 5 m, a wedge angle of 30°, \( g = 9.81 \) m/s², and a friction coefficient of 0.2:
\( \cot30° \approx 1.732 \), so \( 1 – 0.2 \times 1.732 \approx 1 – 0.3464 = 0.6536 \).
Then, \[ v = \sqrt{2 \times 9.81 \times 5 \times 0.6536} \approx \sqrt{64.07} \approx 8.00\, \text{m/s}. \]