Study guide: Net force, acceleration, and work
Use signed quantities to connect net forces to velocity changes and energy.
Ideas to explain in your own words
- Combine opposite forces and distinguish zero net force from zero motion.
- Apply a = Fnet/m with a stated horizontal model.
- Use v = v0 + at when acceleration is constant.
- Use a simple one-dimensional work-energy model.
Watch your reasoning
- With right positive, +8 + (−3) = +5 N, so the net force is 5 N right. Adding the magnitudes without directions would incorrectly give 11 N.
- The net is 12 − 6 = 6 N right; 6/3 = 2 m/s² right.
- 6 + (−2)(2) = +2 m/s. It is still moving right, but slower than initially.
- K = 1/2 × 2 × v² = v². With K = 9 J, v² = 9 and speed is 3 m/s.
A worked review example
Find the net force: Cart A: +6 + (−6) = 0 N. If it is already travelling right at constant speed, these balanced horizontal forces do not change that velocity in the model. Calculate Acceleration from Net Force: Fnet = +6 N. For 2 kg, a = +3 m/s²; for 4 kg, a = +1.5 m/s². Doubling the mass halves acceleration for this unchanged net force. Track Velocity through Slowing and Reversal: At 4 s, v = 6 + (−2)(4) = −2 m/s, left. The cart slows for its first 3 seconds, momentarily stops, then moves left if the acceleration continues. Check Net Work against Kinetic Energy: Wnet = 4 × 4 = 16 J; ΔK = 16 J. For m = 2 kg, K = v² in joules with v in m/s, so v = 4 m/s. Work transfers energy into the cart’s motion.
Make a practice plan
- Try the supplied lesson problems without looking at the answers.
- Compare your method with the worked examples and explain any correction.
- Use the new review cases, then complete the transfer task on paper or in your own document.