A-LevelMathematicsMechanics9709/41 · Oct/Nov 2017 · Q26 Marks
A tractor of mass 3700 kg is travelling along a straight horizontal road at a constant speed of 12 m s⁻¹. The total resistance to motion is 1150 N.
- (i)Find the power output of the tractor's engine.[1]
- (ii)The tractor comes to a hill inclined at 4° above the horizontal. The power output is increased to 25 kW and the resistance to motion is unchanged. Find the deceleration of the tractor at the instant it begins to climb the hill.[3]
- (iii)Find the constant speed that the tractor could maintain on the hill when working at this power.[2]
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2017). The mark or mark type is shown in brackets.
- [B1] Power = 1150 × 12 = 13 800W For use of P = F × v Allow 13.8 kW
- [B1] Driving force = 25000/12 Using F = P/ν
- [M1] 25000/12 – 1150 – 3700g sin 4 = 3700a For applying Newton's 2nd law up the slope, 4 terms
- [A1] a = –0.445 m s⁻²
- [M1] 25000/ν – 1150 – 3700gsin 4 = 0 For stating the equation for constant v, with 3 terms, and solving for v
- [A1] v = 6.70 m s⁻¹
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