Question 10
A construction hoist lifts a 250 kg pallet of bricks vertically through 12 m at a constant speed. The lift takes 2 minutes.
What is the power output of the hoist?
Not quite. You left out $g$ and used $F = 250\text{ N}$: $\dfrac{250 \times 12}{120} = 25\text{ W}$.
That is it. Step through the hints below to see why.
Not quite. You left 2 minutes unconverted: $\dfrac{29\,400}{2} = 14\,700\text{ W} = 14.7\text{ kW}$.
Not quite. You ignored the time and gave the work as the power: $29\,400\text{ W} = 29.4\text{ kW}$.
Orange arrow: the lifting force the option you are testing uses. Orange tick on the time bar: the time that option divides by. Run the lift and watch the power stay the same all the way up.
You used 250 N, the mass in kilograms, as the force. The pallet’s weight is 250 × 9.8 = 2450 N, so your work and power come out 9.8 times too small.
29 400 J of work spread over 120 s is 245 W, at every moment of the lift. This is the power output of the hoist.
Dividing by 2 treats the lift as if it took 2 seconds. The lift takes 2 minutes, which is 120 s.
29 400 J is the work, not the power. Power is the work done each second, so you still divide by the 120 s.
The hoist lifts the pallet at a constant speed, so it pulls up with a force equal to the pallet’s weight.
$$\begin{aligned}W&=Fs\\&=(250\times9.8)\times12\\&=29\,400\text{ J}\end{aligned}$$
At constant speed the lifting force equals the weight, $mg$.
Convert the time to seconds, then divide the work by it.
$$\begin{aligned}P&=\frac{W}{t}\\&=\frac{29\,400}{2\times60}\\&=245\text{ W}\end{aligned}$$
This is option (B).
$W=Fs=(250\times9.8)\times12=29\,400\text{ J}$, and $P=\dfrac{W}{t}=\dfrac{29\,400}{2\times60}=245\text{ W}$.
The past QCAA questions this one was modelled on
Only mark this done when you could do it without help. Reading the solution does not count.
A Tangent Tuition mock exam question. It is not a past QCAA question, and Tangent Tuition is not affiliated with the QCAA.
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