A builder's hoist lifts a bag of cement of mass 40 kg vertically through a height of 15 m in a time of 10 s. Take gravitational field strength g = 10 N/kg. What is the useful power output of the hoist during the lift?

Energy, Work and Power: Question 9

Syllabus 1.7.2, 1.7.4

Multiple choice Core 1 mark

A builder's hoist lifts a bag of cement of mass $40\text{ kg}$ vertically through a height of $15\text{ m}$ in a time of $10\text{ s}$ . Take gravitational field strength $g = 10\text{ N/kg}$ . What is the useful power output of the hoist during the lift?

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Step 1: Recall the equations for work done and power

Lifting at constant speed, the work done equals the gain in gravitational potential energy, $W = mgh$ . Power is the work done per unit time:

$P = \frac{W}{t}$

Step 2: Calculate the work done raising the bag of cement

$W = mgh = 40 \times 10 \times 15$

$W = 6000\text{ J}$

Step 3: Substitute into the power equation

$P = \frac{6000\text{ J}}{10\text{ s}}$

Step 4: Work out the answer

$P = 600\text{ W}$

Why the other options are wrong

Option	How it arises	Error
$60\text{ W}$	$(40 \times 15) \div 10$	Left out $g$ when calculating the work done
$6000\text{ W}$	$40 \times 10 \times 15$	Quoted the work done as the power, forgot to divide by time
$0.0017\text{ W}$	$10 \div 6000$	Formula inverted (time ÷ work done)

Final answers

Useful power output $= \boxed{600\text{ W}}$ , option D