Chapter Eight: Work 8.1 Work 8.2 Efficiency and Power
8.2 Efficiency and Power Every process that is done by machines can be simplified in terms of work: work input: the work or energy supplied to the process (or machine). work output: the work or energy that comes out of the process (or machine).
8.2 Efficiency and Power A rope and pulley machine illustrates a rule that is true for all processes that transform energy. The total energy or work output can never be greater than the total energy or work input.
8.2 Efficiency 65% of the energy in gasoline is converted to heat. As far as moving the car goes, this heat energy is “lost”. The energy doesn’t vanish, it just does not appear as useful output work.
8.2 Efficiency efficiency = Wo Wi The efficiency of a machine is the ratio of usable output work divided by total input work. Efficiency is usually expressed in percent. Output work (J) efficiency = Wo Wi x 100% Input work (J)
Solving Problems You see a newspaper advertisement for a new, highly efficient machine. The machine claims to produce 2,000 joules of output work for every 2,100 joules of input work. What is the efficiency of this machine? Is it as efficient as a bicycle? Do you believe the advertisement’s claim? Why or why not?
Looking for: Given: Relationships: Solution Solving Problems …efficiency of machine Given: …Wi = 2100 J, Wo = 2000 J Relationships: % efficiency = Wo x 100 Wi Solution 2000 J ÷ 2100 J x 100 = 95% efficient
8.2 Power The rate at which work is done is called power. It makes a difference how fast you do work.
8.2 Power Michael and Jim do the same amount of work. Jim’s power is greater because he gets the work done in less time.
8.2 Power Power is calculated in watts. One watt (W) is equal to 1 joule of work per second. James Watt, a Scottish engineer, invented the steam engine. Jame Watt explained power as the number of horses his engine could replace. One horsepower still equals 746 watts.
8.2 Power Work (joules) Power (watts) P = W t Time (s)
Solving Problems Allen lifts his weight (500 newtons) up a staircase that is 5 meters high in 30 seconds. How much power does he use? How does his power compare with a 100-watt light bulb?
Solving Problems Looking for: Given: Relationships: Solution …power Fweight= 500 N; d = 5 m, t = 30 s Relationships: W = F x d; P = W ÷ t Solution W = 500 N x 5 m = 2500 Nm P = 2500 Nm ÷ 30 s = 83 watts Allen’s power is less than a 100-watt light bulb.