AIPMT PRELIMS 2001 Physics Energy Conversion in Mechanical Systems MCQ Question

To find the maximum speed of the child on the swing, we can use the principle of conservation of mechanical energy. The potential energy at the maximum height is converted to kinetic energy at the lowest point. The potential energy (PE) at the maximum height (2 m) can be calculated using the formula: $PE = mgh$, where $m$ is the mass, $g$ is the acceleration due to gravity (approximately $9.81 \, m/s^2$), and $h$ is the height. The maximum height is 2 m, so: $PE = mg(2)$. At the lowest point (0.75 m), the potential energy is: $PE = mg(0.75)$. The change in potential energy is converted into kinetic energy (KE) at the lowest point, given by: $KE = \frac{1}{2} mv^2$. Setting the change in potential energy equal to the kinetic energy gives us: $mg(2 - 0.75) = \frac{1}{2} mv^2$. Simplifying this, we get: $g(1.25) = \frac{1}{2} v^2$ or $v^2 = 2g(1.25)$. Plugging in the value of $g$: $v^2 = 2 \times 9.81 \times 1.25$. Calculating this gives: $v^2 = 24.525$ and thus, $v = \sqrt{24.525} \approx 4.95 \, m/s$. Rounding this to one decimal place gives us approximately 5 m/s, which corresponds to option B. The other options (10 m/s, 8 m/s, and 15 m/s) are incorrect as they exceed the calculated maximum speed based on energy conservation principles.