Power

Work, Energy and Power

(Use g = 9.8 Nkg^-1)

Examples

A Find the work done in lifting a mass of 6 kg through a height of 10 m.

W = Fx W work done in J

F force in N

x distance moved in the direction of the force in m

W = 6 x 9.8 x 10

= 588 J

B) An escalator carries 100 people of average mass 70 kg to a height of 6 m in one minute.

Find the power necessary to do this.

Work done = F x = 100 x (70 x 9.8) x 6 J

Power = rate of doing work = work done per second

= (100 x 70 x 9.8 x 6) / 60 s = 6860 W or 6.86 kW

1. Find the work done to lift a mass of 10 kg through a height of 5 m. If this takes 15s what is the power?

W= Fx = (10kg)(9.8Nkg^-1) (5m) = 490J

P = W/t = 490 J / 15s = 32.67 W

2         A mass of 5 kg is pulled along a surface at a constant speed. Frictional force between the mass and the surface is 4N what work is done in moving the mass 2 m along the surface? Find the power if this takes 2 s.

W = Fx = (4N) (2m) = 8J

P = W/t =  8J/ 2s = 4 W

3        A girl of mass 50 kg runs up a flight of steps 4.5 m high in 5 s. What power does she develop?

W = Fx             F = 50kg x 9.8 = 490.5 N                     x = 4.5m

W = 490.5 x 4.5 = 2207.25 J

P = W/t = 2207.25 J / 5s = 442w

4        2 x 10⁶ kg of water per second flow over a large waterfall of height 50 m.

W = Fx                   F = 2 x 10⁶ kg x 9.8 = 1.96 x10⁺⁷N      x=50m

W= (1.96 x10⁺⁷N) (50m) = 9.8 x10⁺⁸ J

What power is available?

As 1W = I J per second the water falling over the fall transfers 9.8 x10⁺⁸ J of GPE per second the available power is 9.8 x10⁺⁸ W (ignoring losses due to friction)

5        The first-stage rocket motors of Saturn 5 burnt fuel at the rate of 13600 kg per second. The work done by the pumps to drive the fuel into the combustion chambers is the same as that necessary to lift the fuel through 1680 m.

W= Fx                    F = 13600kg x 9.8= 1.3328 x 10⁺⁵ N    x = 1680m

W = (1.3328x 10⁺⁵ ) (1680) = 2.24 x 10⁺⁸ J

What power is needed to do this? (Your answer is about twice the engine power of the largest ocean liner.)

As this mass of fuel is burnt in 1 second then 2.24 x 10⁺⁸ J of work is done in 1 second so the power is 2.24 x 10⁺⁸ W

6        A gas turbine locomotive developing 6330 kW pulls a train at a steady velocity of 108 kmh^{- 1}. What is the total force in Newtons resisting the motion?

At a steady velocity all work done by the engine is against friction or gravity.

The engine does 6.33 x10⁶ J of work every second

As W = Fx the force exerted by the engine is equal Work done per second divided by the distance travelled per second

F = W/x            108 kmh^-1 = (108000m)/(60 min x 60 sec) = 30 ms^-1

F = 6.33 x10⁶ J/ 30m = 211 000N = 2.11 x 10⁵ N

7        A mechanical loader driven by an engine developing 3 kW lifts a total load of 200000 kg to a height of 3 m in one hour. What is the efficiency of the system? 54.4%

Work done lifting 2 x 10⁵kg through 3m

W= Fx              F = (2 x 10⁵kg) (9.8) = 1.96 x 10⁶ N     x = 3m

W = (1.96 x 10⁶ N) (3m) = 5.88 x 10⁷ J

Time = 1hr = 3600 seconds

Power out = (5.88 x 10⁷ J) / (3600) = 1630 W

(Efficiency = Power out / Power in = 1630 / 3000 = 0.54 = 54%)