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**MECHANICS**

**Kinetics**

**Power & Efficiency**

power |

__Power__

Power is by definition __the rate of working__.

Since **work = force x distance moved**, it follows that :

__Example__

A military tank of mass 20 metric tonnes moves up a 30^{o} hill at a uniform speed of 5 ms^{-1} .

If all the frictional forces opposing motion total 5000N, what is the power delivered by the engine?

(g = 10ms^{-2} , answer in kW)

If the tank is moving at constant speed then the forces forwards are balanced by the forces backwards.

m is the tank's mass, then mgsin30^{o} is the component of the weight down the hill

R is the total of resistive forces down the hill

T is the tractive force forwards up the hill

mgsin30^{o} + R = T

T = (20,000 x 10 x 0.5) + 5000 = 105,000N

power = force x speed

power of tank engine = 105,000 x 5 = 525,000W

__Ans. 525 kW__

__Efficiency__

Efficiency is the ratio of useful work out divided by total work done, expressed as a percentage.

__Example__

A pump running at an efficiency of 70% delivers oil at a rate of 4 kgs^{-1} with a speed of 3 ms^{-1}to an oil heater .

If the vertical distance moved by the oil is 10 m, what is the power consumption of the pump?

(g = 10 ms^{-2}, answer to 1 d.p.)

E_{f }= 70%, m=4kg, v=3 ms^{-1} , h=10 m, g=10 ms^{-2}

work/sec. to raise oil 8 m high = mgh = 4x10x10 = 400 J/s

work/sec. to produce discharge speed = 0.5x4x3x3 = 18 J/s

total work/sec. = 400 + 18 = 418 W

418 W represents 70% of the power supplied,

therefore total power consumption of pump =

__Ans. power consumption of pump is 597.1W (1.d.p.)__

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