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Aim: How can we describe power?. Answer Key HW #7. Do Now: A 50 kg box is lifted 10 m in 30 s. How much was was done? W = Fx W = F g h W = mgh W = (50 kg)(10 m/s 2 )(10 m) W = 5,000 J. **No Calculator** 1 min. Power. The rate of doing work Rate means divide by time Scalar quantity.
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Aim: How can we describe power? Answer Key HW #7 Do Now: A 50 kg box is lifted 10 m in 30 s. How much was was done? W = Fx W = Fgh W = mgh W = (50 kg)(10 m/s2)(10 m) W = 5,000 J **No Calculator** 1 min
Power • The rate of doing work • Rate means divide by time • Scalar quantity James Watt 1736-1819 Units:
1. A person pushes a box across a horizontal surface at a constant speed of 0.5 meter per second. The box has a mass of 40 kilograms, and the coefficient of sliding friction is 0.25. The power supplied to the box by the person is • 0.2 W • 5 W • 50 W • 100 W • 200 W **No Calculator** 1 min 15 sec
2. A weight lifter lifts a mass m at constant speed to a height h in time t. What is the average power output of the weight lifter? • mg • mh • mgh • mght • mgh /t **No Calculator** 1 min 15 sec
3. A student weighing 700 N climbs at constant speed to the top of an 8 m vertical rope in 10 s. The average power expended by the student to overcome gravity is most nearly • 1.1 W • 87.5 W • 560 W • 875 W • 5,600 W **No Calculator** 1 min 15 sec
4. Units of power include which of the following? • I. Watt • II. Joule per second • III. Kilowatt‑hour • I only • III only • (C) I and II only • (D) II and III only • (E) I, II, and III I – Watt is the unit of power II – This is W/t (or power) III – This is Pt which equals W **No Calculator** 1 min
5. A child pushes horizontally on a box of mass m which moves with constant speed v across a horizontal floor. The coefficient of friction between the box and the floor is . At what rate does the child do work on the box? • mgv • (B) mgv • (C) mg/v • (D) mg/v • (E) mv2 **No Calculator** 1 min 15 sec