M3 June 2010 Q3

EdexcelOld spec9 marksHooke's Law Work-Energy Principle

Figure 2: particle on a rough inclined plane attached by a spring to O, 1.5 m up the line of greatest slope, plane at angle theta — Figure 2

A particle of mass 0.5 kg is attached to one end of a light elastic spring of natural length 0.9 m and modulus of elasticity \(\lambda\) newtons. The other end of the spring is attached to a fixed point \(O\) on a rough plane which is inclined at an angle \(\theta\) to the horizontal, where \(\sin\theta = \dfrac{3}{5}\). The coefficient of friction between the particle and the plane is 0.15. The particle is held on the plane at a point which is 1.5 m down the line of greatest slope from \(O\), as shown in Figure 2. The particle is released from rest and first comes to rest again after moving 0.7 m up the plane.

Find the value of \(\lambda\). (9)

Force diagram on the inclined plane: normal reaction R, friction F down the plane, weight mg, angle theta — Mark scheme

Scheme	Marks
EPE lost \(= \dfrac{\lambda \times 0.6^2}{2 \times 0.9} - \dfrac{\lambda \times 0.1^2}{2 \times 0.9} \ \left(= \dfrac{7}{36}\lambda\right)\)	M1 A1
R\((\uparrow)\quad R = mg\cos\theta\)	M1
\(= 0.5g \times \dfrac{4}{5} = 0.4g\)
\(F = \mu R = 0.15 \times 0.4g\)	M1 A1
P.E. gained \(=\) E.P.E. lost \(-\) work done against friction \(0.5g \times 0.7\sin\theta = \dfrac{\lambda \times 0.6^2}{2 \times 0.9} - \dfrac{\lambda \times 0.1^2}{2 \times 0.9} - 0.15 \times 0.4g \times 0.7\)	M1 A1 A1
\(0.1944\lambda = 0.5 \times 9.8 \times 0.7 \times \dfrac{3}{5} + 0.15 \times 0.4 \times 9.8 \times 0.7\)
\(\lambda = 12.70\ldots\) \(\lambda = 13\) N or 12.7	A1
	(9 marks)