M2 June 2010 Q2

EdexcelOld spec8 marksWork-Energy Principle

2. A particle \(P\) of mass 0.6 kg is released from rest and slides down a line of greatest slope of a rough plane. The plane is inclined at 30\(^\circ\) to the horizontal. When \(P\) has moved 12 m, its speed is 4 m s\(^{-1}\). Given that friction is the only non-gravitational resistive force acting on \(P\), find

(a) the work done against friction as the speed of \(P\) increases from 0 m s\(^{-1}\) to 4 m s\(^{-1}\), (4)

(b) the coefficient of friction between the particle and the plane. (4)

Diagram: particle sliding 12 m down a plane inclined at 30, from 0 m/s to 4 m/s, with forces R, F and 0.6g — Mark scheme (a)

Scheme	Marks
K.E gained \(= \frac{1}{2} \times 0.6 \times 4^2\)
P.E. lost \(= 0.6 \times g \times (12\sin 30)\)
Change in energy = P.E. lost − K.E. gained
\(= 0.6 \times g \times 12\sin 30 - \frac{1}{2} \times 0.6 \times 4^2\)	M1 A1 A1
\(= 30.48\)
Work done against friction \(= 30\) or 30.5 J	A1
	(4)

Mark scheme (b)

Scheme	Marks
\(\text{R}(\uparrow)\quad R = 0.6g\cos 30\)	B1
\(F = \dfrac{30.48}{12}\)	B1ft
\(F = \mu R\)
\(\mu = \dfrac{30.48}{12 \times 0.6g\cos 30}\)	M1
\(\mu = 0.4987\)
\(\mu = 0.499\ \) or \(\ 0.50\)	A1
	(4)
	(8 marks)