Hooke's law — practice questions

Let the extension be $x$ metres. The particle is in equilibrium, so the forces acting on it are balanced. The upward force is the tension $T$ in the string, and the downward force is the weight of the particle, $mg$.

1. Identify forces: Upward force: Tension, $T$. Downward force: Weight, $W = mg = 2 \times 9.8 = 19.6$ N.
2. Apply equilibrium condition: For equilibrium, the net force is zero. So, $T = W$. $T = 19.6$ N.
3. Apply Hooke's Law: We know $T = \frac{\lambda x}{l_0}$. The given values are $\lambda = 40$ N and $l_0 = 0.8$ m. So, $19.6 = \frac{40 \times x}{0.8}$.
4. Solve for extension $x$: $19.6 = \frac{40}{0.8} x$ $19.6 = 50x$ $x = \frac{19.6}{50} = 0.392$ m.
5. Calculate final length: Final length $L = l_0 + x = 0.8 + 0.392 = 1.192$ m.

Answer: The extension is 0.392 m and the final length is 1.192 m.

The particle is on a smooth horizontal surface, so we only need to consider horizontal forces.

1. Calculate the initial extension: The particle is at B, so the stretched length is $L = 1.8$ m. The natural length is $l_0 = 1.2$ m. Initial extension, $x = L - l_0 = 1.8 - 1.2 = 0.6$ m.
2. Calculate the initial tension: Using Hooke's Law, $T = \frac{\lambda x}{l_0}$. Given $\lambda = 30$ N and $l_0 = 1.2$ m. $T = \frac{30 \times 0.6}{1.2} = \frac{18}{1.2} = 15$ N. This tension acts towards the fixed point A.
3. Apply Newton's Second Law: The only horizontal force acting on the particle at the moment of release is the tension $T$. The net force is therefore $T$. Let the acceleration be $a$. The direction of acceleration is towards A. $F_{net} = ma$ $T = ma$
4. Solve for acceleration $a$: We have $T = 15$ N and $m = 0.5$ kg. $15 = 0.5 \times a$ $a = \frac{15}{0.5} = 30$ m s$^{-2}$.

Answer: The initial acceleration of the particle is 30 m s$^{-2}$ towards A.