A boy uses a single string pulley system to lift a mass of 20kg moving with a velocity of 5ms\(^{-1}\). What is the power developed by the boy if g = 10ms\(...

Power is the rate of doing work, \(P = \dfrac{W}{t}\). When a load is lifted steadily, the work done against gravity in a time \(t\) is \(W = mgh\), and since \(h = vt\) for constant speed, \[P = \frac{mg(vt)}{t} = mgv.\] This is the useful shortcut: the power needed to raise a load at constant velocity is the weight of the load multiplied by the speed of lifting.

Substituting the data, with \(m = 20\,\text{kg}\), \(g = 10\,\text{m s}^{-2}\) and \(v = 5\,\text{m s}^{-1}\): \[P = 20 \times 10 \times 5 = 1000\,\text{W} = 1\,\text{kW}.\]

The phrase "single string pulley system" matters less than it appears. A pulley system can reduce the force the boy must apply, but it cannot reduce the work or the power required, because whatever force is saved is paid for in extra distance pulled. For an ideal system the power developed is still \(mgv\); a real system with friction would need more input power, never less. Note also that \(100\,\text{W}\) comes from omitting \(g\) and computing \(mv\), while \(250\,\text{W}\) comes from using the kinetic-energy expression \(\tfrac{1}{2}mv^{2}\), which is an energy in joules, not a power.

The examination takeaway is to check the unit of the expression you build: \(mgv\) has units \(\text{kg}\cdot\text{m s}^{-2}\cdot\text{m s}^{-1} = \text{J s}^{-1} = \text{W}\), which confirms it is a power before any numbers are substituted.