Section 5–4: Power Physics Coach Kelsoe Pages 179 – 181.

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Section 5–4: Power Physics Coach Kelsoe Pages 179 – 181

Objectives Relate the concepts of energy, time, and power. Calculate power in two different ways. Explain the effect of machines on work and power.

Rate of Energy Transfer The rate at which work is done is called power. Formally, power is defined as is a quantity that measures the rate at which work is done or energy is transformed. The formula for power can be written in two ways: –P = W/Δt P = power, W = work, t = time –P = Fv P = power, F = force, v = velocity

Rate of Energy Transfer Keep in mind that W = Fd, so power then can be solved from the equation P = Fd/Δt. The SI unit of power is the watt, abbreviated with the letter W. It is defined as 1 joule per second. The horsepower, hp, is another unit of power. One horsepower is equal to 746 W.

Sample Problem Power A 193 kg curtain needs to be raised 7.5 m, at constant speed, in as close to 5.0 s as possible. The power ratings for three motors are listed as 1.0 kW, 3.5 kW, and 5.5 kW. Which motor is best for the job?

Sample Problem Solution 1. Identify givens and unknowns –m = 193 kg –Δt = 5.0 s –d = 7.5 m –P = ?

Sample Problem Solution 2. Choose the correct equation –We know that P = W/Δt, but since W = Fd and F = mgd, we can say that P = mgd/Δt

Sample Problem Solution 3. Calculate –P = mgd/Δt –P = (193 kg)(9.81 m/s 2 )(7.5 m)/5.0 s –P = 2800 W or 2.8 kW –So the BEST motor to use would be the 3.5 kW motor because the 5.5 kW motor would lift the curtain too fast and the 1.0 kW motor wouldn’t be fast enough.

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