Problem 22
Question
A mass \(m\) slides down a smooth inclined plane from an initial vertical height \(h,\) making an angle \(\alpha\) with the horizontal. (a) The work done by a force is the sum of the work done by the components of the force. Consider the components of gravity parallel and perpendicular to the surface of the plane. Calculate the work done on the mass by each of the components, and use these results to show that the work done by gravity is exactly the same as if the mass had fallen straight down through the air from a height \(h\) . (b) Use the work-energy theorem to prove that the speed of the mass at the bottom of the incline is the same as if it had been dropped from height \(h,\) independent of the angle \(\alpha\) of the incline. Explain how this speed can be independent of the slope angle. (c) Use the results of part (b) to find the speed of a rock that slides down an icy friction- less hill, starting from rest 15.0 m above the bottom.
Step-by-Step Solution
VerifiedKey Concepts
Gravitational Force Components
- Parallel Component: This part pulls the object down the slope and is given by the formula \( F_{\parallel} = mg \sin \alpha \), where \( m \) is the mass, \( g \) is the acceleration due to gravity, and \( \alpha \) is the angle of the incline.
- Perpendicular Component: This part presses the object into the surface of the incline and does not contribute to the movement down the slope. It is determined by \( F_{\perp} = mg \cos \alpha \).
Inclined Plane Dynamics
- The parallel force component over the incline does work equivalent to \( mgh \). This result matches what would occur if the object fell vertically through the height \( h \). Thus, both methods of descent lead to the same speed at the bottom, since work depends solely on the vertical height and not the path taken.
- This outcome is rooted in the simple relationship \( s = \frac{h}{\sin \alpha} \), which shows how inclined path length \( s \) relates to vertical height \( h \). Substituting this into the work formula, the dependence on \( \alpha \) cancels out.
Conservative Forces
- The total work converted from potential energy (mgh) to kinetic energy is path-independent, as highlighted by the work-energy theorem: \[ \Delta KE = W_{\text{total}} = \frac{1}{2}mv^2 \].
- Since the change in kinetic energy equals the work done by gravity, and this is independent of the angle \( \alpha \), it shows that the speed at the bottom only depends on the height \( h \).