Hint
Choose and retain a positive direction; distinguish signed quantities from magnitudes. Multiply the constant driving force by displacement in its direction.
Worked solution
(a)
The parallel drive does PL, while the opposing resistance does −RL. The other forces do no work.
Answer: 320 J drive; -64 J resistance.
(b)
The work-energy equation is \(4v^2-36=320-64\). It gives \(\displaystyle v^{2}=73\), so the positive speed rounds as stated.
Answer: \(8.54\) m/s.
(c)
The resultant is \(\displaystyle P-R=32 N\), giving \(\displaystyle a= \frac{4m}{s^{2}}\). Then \(v^2=9+2(4)(8)=73\) recovers the same speed.
Answer: \(8.54\) m/s.
Maths4U indicative marking guidance
Maths4U indicative marking guidance, not an official mark scheme. Credit complete valid alternative methods and consequential reasoning. An unexplained result does not earn a method mark.
(a) · 3 marks
- 1 mark: Use force times displacement for each named constant force.
- 1 mark: Retain the negative sign for work by the opposing resistance.
- 1 mark: Give the complete requested result, using the required units, precision and boundary inclusion.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: Add signed work by every contributing force.
(b) · 4 marks
- 1 mark: Calculate the net signed work.
- 1 mark: Equate it to the kinetic-energy change, retaining the initial energy.
- 1 mark: Solve for the positive speed and round only the final result.
- 1 mark: Give the complete requested result, using the required units, precision and boundary inclusion.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: Keep the initial kinetic energy when the particle does not start from rest.
(c) · 3 marks
- 1 mark: Use Newton’s second law to find the constant acceleration.
- 1 mark: Use the time-free kinematics equation with the original initial speed and travel distance.
- 1 mark: Give the complete requested result, using the required units, precision and boundary inclusion.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: Add signed work by every contributing force.
Total: 10 marks.