\includegraphics{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 \(\sin \theta = \frac{3}{5}\). 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]

EPE lost $= \frac{\lambda \times 0.6^2}{2 \times 0.9} - \frac{\lambda \times 0.1^2}{2 \times 0.9} = \frac{7}{36}\lambda$ | M1 A1 |
R (↑): $R = mg\cos\theta = 0.5g \times \frac{4}{5} = 0.4g$ | M1 |
$F = \mu R = 0.15 \times 0.4g$ | M1 A1 |
P.E. gained $=$ E.P.E. lost $-$ work done against friction | |
$0.5g \times 0.7\sin\theta = \frac{\lambda \times 0.6^2}{2 \times 0.9} - \frac{\lambda \times 0.1^2}{2 \times 0.9} - 0.15 \times 0.4g \times 0.7$ | M1 A1 A1 |
$0.1944\lambda = 0.5 \times 9.8 \times 0.7 \times \frac{3}{5} + 0.15 \times 0.4 \times 9.8 \times 0.7$ | |
$\lambda = 12.70\ldots = 13 \text{ N or } 12.7$ | A1 | (9 marks total)

\includegraphics{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 $\sin \theta = \frac{3}{5}$. 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]

\hfill \mbox{\textit{Edexcel M3  Q3 [9]}}