PHY183-Lecture15pre - Power for a Constant Force Physics for Scientists Engineers 1 Final answer Or Physics for Scientists&Engineers 1 1 February 8

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1 February 8, 2012 Physics for Scientists&Engineers 1 1 Physics for Scientists & Engineers 1 Spring Semester 2012 Lecture 15 Power and Potential Energy February 8, 2012 Physics for Scientists&Engineers 1 2 Power for a Constant Force Work for a constant force: Differential: (because F =constant) According to definition of power: Final answer: Or: W = F • Δ r dW = F d r P = dW dt = F d r dt P = F v P = Fv cos α Fv February 8, 2012 Physics for Scientists&Engineers 1 3 Accelerating Car Question: Car of mass 3,120 lb (= 1,415 kg) can reach the 60 mph (= 26.8 m/s) mark in 4.0 seconds What is the average power needed to accomplish this? Answer: Car’s kinetic energy at 60 mph Work-kinetic energy theorem: K = 1 2 mv 2 = 1 2 (1415 kg) (26.8 m/s) 2 = 508 kJ W = Δ K = K K 0 = 508 kJ February 8, 2012 Physics for Scientists&Engineers 1 4 Accelerating Car Average power to get to the 60 mph mark in 4.0 seconds: Discussion: Is this realistic? No! 170 hp are not enough to accomplish this acceleration for a car of mass 1415 kg • This Corvette has 430 hp Why? Not perfectly efficient engine; not able to produce peak hp at all rpm; energy loss from friction and drag Done P = W Δ t = 5.08 10 5 J 4.0 s = 127 kW = 170 hp February 8, 2012 Physics for Scientists&Engineers 1 5 Power to Keep a Car Moving A car travels at 60.0 mph on a level road. The car has a drag coefficient of 0.33 and a frontal area of 2.2 m 2 . How much power does the car need to maintain its speed? Take the density of air to be 1.29 kg/m 3 . We can write the power as The force here is the force of air resistance We can combine these equations to get So the power required to overcome air resistance is P = Fv = F drag v F drag = Kv 2 K = 1 2 c d A ρ air P = 1 2 c d A air v 2 v = 1 2 c d A air v 3 P = 1 2 0.33 () 2.2 m 2 1.29 kg/m 3 26.8 m/s 3 = 9.01 kW = 12.1 hp Lifting Bricks PROBLEM A load of bricks at a construction site has a mass of 85.0 kg. A crane raises this load from the ground to a height of 50.0 m in 60.0 s at a low constant speed. What is the average power of the crane? THINK Raising the bricks at a low constant speed means that the kinetic energy is negligible, so the work in this situation is done against gravity only There is no acceleration, and friction is negligible The average power then is just the work done against gravity divided by the time it takes to raise the load of bricks to the stated height SKETCH In the free body diagram, the tension on the cable is equal to the weight of the bricks The load is moved vertically a distance h 2/8/2012 Physics for Scientists&Engineers 1 6
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2 Lifting Bricks RESEARCH The work done by the crane is The average power required to lift the load in the given time is SIMPLIFY CALCULATE 2/8/2012 Physics for Scientists&Engineers 1 7 W = mgh P = W Δ t P = mgh Δ t P = 85.0 kg () 9.81 m/s 2 50.0 m 60.0 s = 694.875 W Lifting Bricks ROUND DOUBLE-CHECK Convert out result for the required average power from watts to horsepower So a 1 hp motor would be sufficient to lift the 85.0 kg
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This note was uploaded on 04/02/2012 for the course PHY 183 taught by Professor Wolf during the Spring '08 term at Michigan State University.

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PHY183-Lecture15pre - Power for a Constant Force Physics for Scientists Engineers 1 Final answer Or Physics for Scientists&Engineers 1 1 February 8

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