A pendulum with a bob of mass 5 kg is held stationary at a height h metres above the ground - NSC Physical Sciences - Question 5 - 2016 - Paper 1

To find the height h, we can use the principle of conservation of mechanical energy (before and after the collision). The mechanical energy at the height before the collision should equal the sum of kinetic and potential energy after the collision:

Before collision: $PE_{initial} = mgh \\ (5)(9.8)h$

After collision: $E_k + PE_{final} = E_k + mgh_{final} \\ = 24.50 + (5)(9.8)(\frac{h}{4})$ Setting both mechanical energies equal: $(5)(9.8)h = 24.50 + (5)(9.8)(\frac{h}{4})$

By solving this equation, we can find the value of h, which is approximately $0.67 \, m$.

The work-energy theorem states that the net work done on an object is equal to the change in the object's kinetic energy.

The work done by the frictional force can be calculated by considering the change in mechanical energy between points B and C. The formula for work is: $W = \Delta KE + \Delta PE$ Where:

• $\,\Delta KE = (KE_b - KE_c)$\
• $\,\Delta PE = (PE_C - PE_B)$

Given:

• $KE_b = \frac{1}{2} mv^2 = \frac{1}{2} (2) (4.95)^2 = 24.50 \, J$
• $KE_c = \frac{1}{2} (2) (2)^2 = 4 \, J$
• $\Delta KE = 24.50 - 4 = 20.50 \, J$

Now considering the change in potential energy:

• Since going from B to C the block rises $0.5 \, m$, using $PE = mgh$:
• $PE_C - PE_B = (2)(9.8)(0.5) = 9.8 \, J$

Thus, the work done by the friction force is: $W = \Delta KE + \Delta PE = 20.50 - 9.8 = 10.70 \, J$