Section 3.5: Conservation of Energy
The principle of conservation of energy states that the total mechanical energy of an isolated system remains constant if only conservative forces act. Energy can be transformed between kinetic and potential forms without loss.
\( E_\text{mech} = K + U \)
- \( K = \frac{1}{2}mv^2 \): Kinetic Energy
- \( U = mgh \): Gravitational Potential Energy
\( E_\text{mech,initial} = E_\text{mech,final} \)
Or equivalently, \( K_i + U_i = K_f + U_f \)
Example 1
A 2 kg object is dropped from a height of 5 m. Find its speed just before hitting the ground using conservation of energy.
\( U_i = mgh = 2 \cdot 9.8 \cdot 5 = 98 \, \text{J} \)
\( K_f = E_\text{mech} - U_f = 98 - 0 = 98 \, \text{J} \)
\( \frac{1}{2} m v^2 = 98 \Rightarrow v = \sqrt{\frac{2 \cdot 98}{2}} = \sqrt{98} \approx 9.9 \, \text{m/s} \)
Example 2
A roller coaster car of mass 500 kg starts from rest at 20 m height. Find its speed at 5 m height.
\( K_i + U_i = 0 + 500 \cdot 9.8 \cdot 20 = 98,000 \, \text{J} \)
\( U_f = mgh = 500 \cdot 9.8 \cdot 5 = 24,500 \, \text{J} \)
\( K_f = 98,000 - 24,500 = 73,500 \, \text{J} \)
\( v = \sqrt{\frac{2 K_f}{m}} = \sqrt{\frac{2 \cdot 73,500}{500}} \approx 17.1 \, \text{m/s} \)
Practice Problems
- A 3 kg object falls from 10 m. Compute speed just before impact.
- A 5 kg mass slides down a frictionless ramp from 8 m. Find final speed.
- A ball of mass 2 kg is thrown vertically with 15 m/s. Find maximum height using energy conservation.
- A roller coaster of mass 400 kg goes from 25 m to 10 m height. Determine speed at lower point.
- An object of mass 6 kg is dropped from 12 m. Find kinetic energy just before hitting ground.