A car of mass 1200 kg travels at a constant speed of 15 m s$^{-1}$ up a straight road inclined at 4.0° to the horizontal. The total resistive force on the car is 450 N.
(a) Calculate the useful power output of the car's engine. [4]
(b) The engine's input power is 64 kW. Calculate its efficiency. [1]
- (a) component of weight down the slope = $1200 \times 9.81 \times \sin4.0° = 821$ N(1)
- (a) driving force = $821 + 450 = 1270$ N (constant speed so resultant force zero)(1)
- (a) $P = Fv = 1271 \times 15$(1)
- (a) $1.9\times10^{4}$ W allow 19 kW(1)
- (b) efficiency = $\dfrac{19.1}{64} = 0.30$ / 30% ecf(1)
- Answer: (a) $1.9\times10^{4}$ W (b) 30%
- Examiner tip: At constant speed the driving force equals the TOTAL opposing force — include the component of weight down the slope, not just friction.