Section 4.4: Energy and Inclined Surfaces

When a block moves on an inclined surface, its gravitational potential energy and kinetic energy change depending on height and speed. Inclined surfaces affect acceleration and energy conversion.

Gravitational Potential Energy: \[ PE = m g h \] where \(h = L \sin \theta\), \(L\) is the distance along incline, and \(\theta\) is incline angle.
Kinetic Energy: \[ KE = \frac{1}{2} m v^2 \]
Energy Conservation on Incline: \[ m g h = \frac{1}{2} m v^2 \quad \Rightarrow \quad v = \sqrt{2 g h} \]

If friction exists, subtract work done by friction: \( W_f = f_k L = \mu_k m g \cos \theta \cdot L \)

Example 1

A 2 kg block slides down a frictionless incline of 30° and length 5 m. Find its speed at the bottom.

Height: \( h = 5 \sin 30^\circ = 2.5 \, m \)

\( v = \sqrt{2 g h} = \sqrt{2 \cdot 9.8 \cdot 2.5} \approx 7 \, \text{m/s} \)

Example 2

A 3 kg block slides down a 4 m incline with \(\mu_k = 0.1\). Incline angle 20°. Find speed at bottom.

Height: \( h = 4 \sin 20^\circ \approx 1.37 \, m \)

Friction work: \( W_f = \mu m g \cos \theta \cdot L = 0.1 \cdot 3 \cdot 9.8 \cdot \cos 20^\circ \cdot 4 \approx 11.5 \, J \)

Initial PE: \( m g h = 3 \cdot 9.8 \cdot 1.37 \approx 40.2 \, J \)

KE at bottom: \( 40.2 - 11.5 = 28.7 \Rightarrow v = \sqrt{2 \cdot 28.7 / 3} \approx 4.37 \, m/s \)

Practice Problems

  1. A 5 kg block slides down a frictionless 10 m incline at 25°. Find speed at bottom.
  2. Inclined plane, 3 m long, \(\mu_k = 0.2\), block mass 2 kg, angle 30°. Find speed at base.
  3. Block starts from rest at height 4 m. Incline angle 15°, frictionless. Determine kinetic energy at bottom.
  4. Incline length 6 m, angle 20°, block mass 1.5 kg, \(\mu_k = 0.05\). Compute work done by friction.
  5. A block slides down incline. Initial PE 50 J, friction 10 J. Find KE at bottom.
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