Hint
Choose and retain a positive direction; distinguish signed quantities from magnitudes. Convert instantaneous power to driving force at the specified nonzero speed.
Worked solution
(a)
Convert instantaneous power to driving force at the specified nonzero speed.
The power relation gives \(\displaystyle F= \frac{P}{v}=\)\(1454\) N.
Answer: \(1454\) N
(b)
Use Newton’s second law with every along-slope force.
Subtract resistance 54 and gravity component \(1200\) N. Dividing the remaining 200 N by mass gives \(\displaystyle \frac{1m}{s^{2}}\).
Answer: \(1\) \(\displaystyle \frac{m}{s^{2}}\)
(c)
Set acceleration to zero only for the explicitly changed steady-motion condition.
At steady speed \(\displaystyle F=R+Mg \sin \alpha\), giving power \(8778\) W.
Answer: \(8778\) W
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) · 2 marks
- 1 mark: Convert instantaneous power to driving force at the specified nonzero speed.
- 1 mark: Give the correct requested result, retaining signs, units and any domain restrictions.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: Subtract both gravity’s downslope component and the stated resistance.
(b) · 2 marks
- 1 mark: Use Newton’s second law with every along-slope force.
- 1 mark: Give the correct requested result, retaining signs, units and any domain restrictions.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: Do not treat the instantaneous acceleration of a constant-power model as a constant over time.
(c) · 2 marks
- 1 mark: Set acceleration to zero only for the explicitly changed steady-motion condition.
- 1 mark: Give the correct requested result, retaining signs, units and any domain restrictions.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: Subtract both gravity’s downslope component and the stated resistance.
Total: 6 marks.