M2 June 2010 Q4

EdexcelOld spec8 marksPower

4. A car of mass 750 kg is moving up a straight road inclined at an angle \(\theta\) to the horizontal, where \(\sin\theta = \dfrac{1}{15}\). The resistance to motion of the car from non-gravitational forces has constant magnitude \(R\) newtons. The power developed by the car’s engine is 15 kW and the car is moving at a constant speed of 20 m s\(^{-1}\).

(a) Show that \(R = 260\). (4)

The power developed by the car’s engine is now increased to 18 kW. The magnitude of the resistance to motion from non-gravitational forces remains at 260 N. At the instant when the car is moving up the road at 20 m s\(^{-1}\) the car’s acceleration is \(a\) m s\(^{-2}\).

(b) Find the value of \(a\). (4)

Diagram: car on a slope at angle theta moving up at 20 m/s with acceleration a, tractive force T prime up the slope, resistance 260 N down the slope and weight 750g — Mark scheme (b)

Scheme	Marks
\(T = \dfrac{15000}{20} = 750\)	M1
R(parallel to road) \(\ T = R + 750g\sin\theta\)	M1 A1
\(R = 750 - 750 \times 9.8 \times \frac{1}{15}\)
\(R = 260\ *\)	A1
	(4)

Scheme	Marks
\(T^{\prime} = \dfrac{18000}{20} = 900\)	M1
\(T^{\prime} - 260 - 750g \times \sin\theta = 750a\)	M1 A1
\(a = \dfrac{900 - 260 - 750 \times 9.8 \times \frac{1}{15}}{750}\)
\(a = 0.2\)	A1
	(4)
	(8 marks)