iGCSE Physics (0625) 1.7.2 Work -Exam Style Questions Paper 2 - New Syllabus
▶️ Answer/Explanation
Detailed solution:
By the principle of conservation of energy (or work done), the work done by the father equals the total work done on both boys.
Father’s work: \(W = F \times d = 1000\text{ N} \times 0.40\text{ m} = 400\text{ J}\).
Total work on boys: \((W_b \times 0.20) + (W_b \times 0.80) = W_b \times 1.00\text{ m}\).
Equating: \(W_b \times 1.00 = 400 \Rightarrow W_b = 400\text{ N}\).
This matches Option B.
▶️ Answer/Explanation
Detailed solution:
Initial gravitational potential energy $= \text{weight} \times \text{height} = 20\text{ N} \times 2.0\text{ m} = 40\text{ J}$.
Kinetic energy at bottom $= \text{Initial GPE} – \text{Work done against friction} = 40\text{ J} – 10\text{ J} = 30\text{ J}$.
Mass of box $= \frac{\text{weight}}{g} = \frac{20\text{ N}}{10\text{ m/s}^2} = 2.0\text{ kg}$.
Using $\frac{1}{2}mv^2 = 30\text{ J}$, $v^2 = \frac{2 \times 30}{2.0} = 30$, so $v = \sqrt{30} \approx 5.4\text{ m/s}$.
This matches Option A.
▶️ Answer/Explanation
Detailed solution:
The work done by a constant force is the product of the force and the distance moved in the direction of the force.
Here, the tension \(F\) acts along the slope, and the object moves distance \(d\) along the slope.
Therefore, the work done by the boy is simply \(W = F \times d\).
The vertical height \(h\) is relevant for work done against gravity, not the work done by the rope’s tension.
Thus, Option A is correct.