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 1kW = 1000W

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

Find the magnitude of the resistance forces at this speed Example 1 A cyclist travels at a constant speed of 10ms-1 along a straight horizontal road. The power the cyclist develops is 400W and the mass of the cyclist and bicycle totals 80 kg. Find the magnitude of the resistance forces at this speed The resistance forces are found to be proportional to the cyclist’s speed b Write down the equation for the resistance forces 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 10 ms-1 R F 80g

Now summarise the information P = 400 W F = ? v = 10 ms-1 a. Apply P = Fv 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 = 40N 40 = k x 10 k = 4 R = 4v

Apply F = ma remember this F is the resultant force 200 – 8 = 80a normal reaction 2 ms-1 R F 80g R = ? V = 2 ms-1 R = 4v R = 4 x 2 R = 8N P = Fv 400 = F x 2 F = 200 N Apply F = ma remember this F is the resultant force 200 – 8 = 80a 80a = 192 a = 2.4 ms-2

The work done by the pump The power developed by the pump Example 2 A pump raises 20 kg of water to a height of 5m in 30 seconds. The water starts at rest and it is issued at 5ms-1. Calculate: The work done by the pump The power developed by the pump State one key assumption you have made. 5 ms-1 20 kg delivered in 30 seconds 0 ms-1 5 m pump zero PE level The work done by the pump is the same as the increase in mechanical energy. W = KEend + PEend W = mgh + ½mv² W = (20 x 9.8 x 5) + (½ x 20 x 5²) W = 1230 J

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

Example 3. A car, of mass 1000kg, produces 50 kW of power. The resistances to motion are given by R = 100v 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 = 100v 1000g 10º Use P = Fv F = Resolve forces parallel to the plane F – R – 1000gsin10º = 0 - 1000v – 1701.8 = 0 50000 – 1000v² - 1701.8v = 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 enjoy Mechanics 2 pg 92 exercise 4c all questions