Power Key ideas Power is the rate at

Power

Key ideas • Power is the rate at which work is done • Power = Force x velocity Power = Fv and Power = work done ÷ time taken P= Power is measured in watts (W) You will probably use kilowatts 1 k. W = 1000 W

How to answer the questions 1. Draw a clear diagram 2. Summarise the information given 3. It may be necessary to calculate the work done using KE = ½mv², 4. PE = mgh and W = Fs 4. Use P = Fv or P = 5. You may also need to resolve forces or to apply F = ma where F is the resultant force

Example 1 A cyclist travels at a constant speed of 10 ms-1 along a straight horizontal road. The power the cyclist develops is 400 W and the mass of the cyclist and bicycle totals 80 kg. a. Find the magnitude of the resistance forces at this speed b. The resistance forces are found to be proportional to the cyclist’s speed c. b Write down the equation for the resistance forces d. C Calculate the acceleration of the cyclist when his speed is 2 ms-1 if the power developed remains unchanged Draw a clear diagram then click to continue normal reaction R 10 ms-1 F 80 g

Now summarise the information P = 400 W a. F=? v = 10 ms -1 Apply P = Fv 400 = F x 10 F = 40 N b. R v So R = kv k is the constant of proportion because the speed is constant R = 40 N 40 = k x 10 k=4 R = 4 v

c. a normal reaction R 2 ms-1 F 80 g P = Fv R=? 400 = F x 2 R = 4 v F = 200 N R=4 x 2 V = 2 ms-1 R = 8 N Apply F = ma 200 – 8 = 80 a = 192 a = 2. 4 ms-2 remember this F is the resultant force

Example 2 A pump raises 20 kg of water to a height of 5 m in 30 seconds. The water starts at rest and it is issued at 5 ms-1. Calculate: a. The work done by the pump b. The power developed by the pump c. State one key assumption you have made. 5 ms-1 0 ms-1 20 kg delivered in 30 seconds 5 m pump zero PE level a. The work done by the pump is the same as the increase in mechanical energy. b. W = KEend + PEend c. W = mgh + ½mv² 1. W = (20 x 9. 8 x 5) + (½ x 20 x 5²) 2. W = 1230 J

5 ms-1 0 ms-1 20 kg delivered in 30 seconds 5 m pump b. P = ? W = 1230 J c. P= d. P= e. P = 41 W zero PE level t = 30 s f. We assume the force produced by the pump is constant

Example 3. A car, of mass 1000 kg, produces 50 k. W of power. The resistances to motion are given by R = 100 v where v is the speed of the car. Find the maximum speed the car can attain ascending (going up) a road inclined at 10º to the horizontal. V ms-1 Normal reaction F P = 50 000 W R = 100 v 10º Use P = Fv F= 1000 g Resolve forces parallel to the plane F – R – 1000 gsin 10º = 0 - 1000 v – 1701. 8 = 0 50000 – 1000 v² - 1701. 8 v = 0 Use the quadratic formula to solve v = 15. 4 ms-1 (3 s. f. )

Try some yourself Blue book pg 267, 268, 269, 270 Mechanics 2 pg 92 exercise 4 c all questions enjoy