Section 3.4: Kinetic Energy & Work-Energy Theorem
Kinetic energy represents the energy associated with the motion of an object. The work-energy theorem relates the net work done on an object to its change in kinetic energy.
Kinetic Energy:
\( K = \frac{1}{2} m v^2 \)
- \( m \): Mass of the object
- \( v \): Speed of the object
Work-Energy Theorem:
\( W_\text{net} = \Delta K = K_f - K_i \)
- \( W_\text{net} \): Net work done on the object
- \( K_i, K_f \): Initial and final kinetic energies
Example 1
A 2 kg object accelerates from 3 m/s to 7 m/s. Find the net work done on the object.
\( K_i = \frac{1}{2} \cdot 2 \cdot 3^2 = 9 \, \text{J} \)
\( K_f = \frac{1}{2} \cdot 2 \cdot 7^2 = 49 \, \text{J} \)
\( W_\text{net} = K_f - K_i = 49 - 9 = 40 \, \text{J} \)
Example 2
A car of mass 1000 kg accelerates from 0 to 20 m/s. Calculate the net work done.
\( K_i = 0 \)
\( K_f = \frac{1}{2} \cdot 1000 \cdot 20^2 = 200,000 \, \text{J} \)
\( W_\text{net} = K_f - K_i = 200,000 \, \text{J} \)
Practice Problems
- An object of mass 5 kg accelerates from 2 m/s to 6 m/s. Find the net work done.
- A 10 kg object is moving at 3 m/s. Compute its kinetic energy.
- A car of mass 800 kg slows from 25 m/s to 10 m/s. Find the work done by braking.
- A 15 kg box accelerates from rest to 4 m/s. Find the work done.
- An object of 2 kg accelerates from 5 m/s to 10 m/s. Compute net work done.