Work and Energy Legacy Problem #6 Guided Solution
Problem*
During the Powerhouse lab, Jerome runs up the stairs, elevating his 102 kg body a vertical distance of 2.29 meters in a time of 1.32 seconds at a constant speed.
- Determine the work done by Jerome in climbing the stair case.
- Determine the power generated by Jerome.
Audio Guided Solution
An effective problem solver reads the problem carefully and generates a mental picture of what's going on, identifying the known and the unknown quantities, and then plots a strategy as to how to get from known to unknown quantities. Here what we have is a picture of Jerome running up the stairs during a lap. The purpose of the lap is ultimately to determine his power. So as Jerome runs up the stairs, what we know is there must be enough up force to lift his body at a constant speed. If his mass is 102 kilograms, we should be able to calculate the down force of gravity, and the up force has to balance this in order to have a constant speed motion. We know the distance at which Jerome moves vertically up the stairs, and so presuming an up force to lift his body up, what we could do is calculate the work done using Fd cos theta. Ultimately we wish to calculate the power in Part B. The power, the rate at which work is done, is simply the ratio of work to time. So we'll take our answer in Part A and divide by time. That will give us our power. That's our strategy. Let's employ it. In order to calculate the force up, we have to determine the force of gravity down upon Jerome. That force of gravity is just m times g, where the m is 102 and the g is 9.8 newtons per kilogram. We would get 999.60 newtons. Now to find the work, we have to use this force value as the up value. We'd go 999.60 newtons multiplied by the 2.29 meters that Jerome elevates his body upwards, and we would get a work done of 2289.08 joules. We could round that for Part A to 2.30 times 10 to the third joules. Now to determine the power, we need to take the ratio of work to time. We need to take this 2289.08 joules and divide it by 1.32 seconds. When we do that, we get 1.734 times 10 to the third joules. To three digits, it would be 1.73 times 10 to the third. And what is the unit of power? I didn't mean to ask that. I meant to say that. What is the unit of power? That sounds like a question again. What I'm trying to say is, what? The unit what? That's the unit of power. That's what I meant. I'm only trying to ask the question, what is the unit of power? Oh, never mind.
Solution
- 2.30 x 103 J
- 1.73 x 103 W
Habbits of an Effective Problem Solver
- Read the problem carefully and develop a mental picture of the physical situation. If necessary, sketch a simple diagram of the physical situation to help you visualize it.
- Identify the known and unknown quantities in an organized manner. Equate given values to the symbols used to represent the corresponding quantity - e.g., \(\descriptive{v}{v,velocity}_\descriptive{o}{o,original} = 0 \unit{\meter\per\second}\); \(\descriptive{a}{a,acceleration} = 4.2\unit{\meter\per\square\second}\); \(\descriptive{v}{v,velocity}_\descriptive{f}{f,final} = 22.9 \unit{\meter\per\second}\); \(\descriptive{d}{d,distance} = \colorbox{gray}{Unknown}\).
- Use physics formulas and conceptual reasoning to plot a strategy for solving for the unknown quantity.
- Identify the appropriate formula(s) to use.
- Perform substitutions and algebraic manipulations in order to solve for the unknown quantity.
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