THE PRACTICE STUDIO
Go slowly. Ask why. Make it yours.
PRACTICE / Power
Power needed to raise the load
A hoist lifts 30 kg vertically through 8 m in 6 s at constant speed.
Find its useful mechanical power.
Organize the givens & assumptions
- m = 30 kg
- h = 8 m
- t = 6 s
g = 10 m/s²; constant lifting speed; useful output is the gravitational energy increase of the load.
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Problem: Power needed to raise the load A hoist lifts 30 kg vertically through 8 m in 6 s at constant speed. Question: Find its useful mechanical power. Givens: - m = 30 kg - h = 8 m - t = 6 s Assumptions: g = 10 m/s²; constant lifting speed; useful output is the gravitational energy increase of the load. Choices (LaTeX): A. 50\ \mathrm W B. 40\ \mathrm W C. 400\ \mathrm W D. 2400\ \mathrm W Ask for my approach first. Help me distinguish a physical misconception from an algebra error without revealing which choice is correct. Tutoring approach: Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.
Start with the smallest hint.
Work through it with me.
One decision at a time. Pause, predict, then unpack the algebra.
At constant speed the useful work is the increase in gravitational potential energy.
Choose a physical relation
Constant speed leaves kinetic energy unchanged, so the useful work raises gravitational potential energy.
Isolate what is asked
At constant speed the useful work is the increase in gravitational potential energy.
Why is this step valid?
Divide the 2400 J energy gain by the 6 s lifting time to obtain watts.
Substitute and interpret
Useful power is 400 W; electrical input could be greater if the hoist has losses.
Force times lifting speed gives the same power as work per time.
TRY THE IDEA AGAIN
Can electrical input equal 300 W for this steady lift without another energy source?
YOUR LEARNING NOTE
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