Calculate the work done in raising the pendulum to point B, if the mass of the bob is 50 g. [g = 10 m s⁻²]

The work done in raising the pendulum is equal to the change in its gravitational potential energy, $W = mgh$.
Given:
Mass of bob, $m = 50 \text{ g} = 0.050 \text{ kg}$
Acceleration due to gravity, $g = 10 \text{ m/s}^2$
Length of the pendulum, $L = 60 \text{ cm} = 0.60 \text{ m}$
Angle of displacement, $\theta = 60^\circ$

The vertical height $h$ is given by the formula $h = L(1 - \cos \theta)$.

Note: The "10 m" label in the diagram is inconsistent with the pendulum length and angle, so it is disregarded.

Hooke's Law states that the force exerted by a spring is proportional to its extension, $F = kx$.
The force is the weight of the mass, $F = Mg$.
The extension $x$ is the total length of the spring ($L_{total}$) minus its natural length ($l$), so $x = L_{total} - l$.
Thus, $Mg = k(L_{total} - l)$.
The mass of the scale pan is $m_{pan} = 0.04 \text{ kg}$.
Acceleration due to gravity, $g = 10 \text{ m/s}^2$.

Case 1:
Object mass $m_1 = 0.20 \text{ kg}$.
Total mass $M_1 = m_{pan} + m_1 = 0.04 \text{ kg} + 0.20 \text{ kg} = 0.24 \text{ kg}$.
Length of the spring $L_1 = 0.055 \text{ m}$.
Equation 1: $M_1 g = k(L_1 - l)$

Case 2:
Object mass $m_2 = 0.28 \text{ kg}$.
Total mass $M_2 = m_{pan} + m_2 = 0.04 \text{ kg} + 0.28 \text{ kg} = 0.32 \text{ kg}$.
Length of the spring $L_2 = 0.065 \text{ m}$.
Equation 2: $M_2 g = k(L_2 - l)$

From equation (1), $k = \frac{2.4}{0.055 - l}$.
Substitute this expression for $k$ into equation (2):

Using equation (1):