| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Elastic string equilibrium and statics |
| Difficulty | Standard +0.3 This is a straightforward M3 equilibrium problem requiring resolution of forces in two directions to find tension, then applying Hooke's law to find extension and using the standard EPE formula. All steps are routine applications of standard methods with no novel insight required, making it slightly easier than average. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| resolve \(\rightarrow\): \(20 - T\sin 30° = 0\) \(\therefore T = 40\) N | M1 A2 | |
| let natural length be \(l\) \(\therefore T = \frac{\lambda l}{l} = \frac{80l(2-l)}{l}\) | M1 A1 | |
| \(T = 40 \therefore l = 2(1.2 - l)\) giving \(l = 0.8\) m | A1 | |
| EPE \(= \frac{\lambda x^2}{2l} = \frac{80 \times 0.4^2}{2 \times 0.8} = 8\) J | M1 A1 | (8) |
| resolve $\rightarrow$: $20 - T\sin 30° = 0$ $\therefore T = 40$ N | M1 A2 | |
| let natural length be $l$ $\therefore T = \frac{\lambda l}{l} = \frac{80l(2-l)}{l}$ | M1 A1 | |
| $T = 40 \therefore l = 2(1.2 - l)$ giving $l = 0.8$ m | A1 | |
| EPE $= \frac{\lambda x^2}{2l} = \frac{80 \times 0.4^2}{2 \times 0.8} = 8$ J | M1 A1 | (8) |2. A particle $P$ is attached to one end of a light elastic string of modulus of elasticity 80 N . The other end of the string is attached to a fixed point $A$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{ad523c3f-9109-45a8-8399-80a4c2edeff7-2_410_570_1210_735}
\captionsetup{labelformat=empty}
\caption{Fig. 1}
\end{center}
\end{figure}
When a horizontal force of magnitude 20 N is applied to $P$, it rests in equilibrium with the string making an angle of $30 ^ { \circ }$ with the vertical and $A P = 1.2 \mathrm {~m}$ as shown in Figure 1.
\begin{enumerate}[label=(\alph*)]
\item Find the tension in the string.
\item Find the elastic potential energy stored in the string.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 Q2 [8]}}