A-LevelMathematicsMechanics9709/43 · Oct/Nov 2017 · Q26 Marks
A lorry of mass 7850 kg travels on a straight hill which is inclined at an angle of 3° to the horizontal. There is a constant resistance to motion of 1480 N.
- (i)Find the power of the lorry's engine when the lorry is going up the hill at a constant speed of 10 m s⁻¹.[3]
- (ii)Find the power of the lorry's engine at an instant when the lorry is going down the hill at a speed of 15 m s⁻¹ with an acceleration of 0.8 m s⁻².[3]
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2017). The mark or mark type is shown in brackets.
- [F = 1480 + 7850g sin 3] ( = 5588) M1
- [P / 10 = 1480 + 7850g sin 3] →P = ... M1 Using P = Fv and solving for P
- Power = 55 900 W A1
- [F+7850g sin 3 – 1480 = 7850 × 0.8] (F = 3652) M1 Use of Newton's Second Law
- [P / 15 + 7850g sin 3 – 1480 = 7850 × 0.8] →P = ... M1 Using P = Fv and solving for P
- Power = 54800 W A1
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