Power is by definition the rate of working.
Since work = force x distance moved, it follows that :
A military tank of mass 20 metric tonnes moves up a 30o 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 mgsin30o 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
mgsin30o + 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 is the ratio of useful work out divided by total work done, expressed as a percentage.
A pump running at an efficiency of 70% delivers oil at a rate of 4 kgs-1 with a speed of 3 ms-1to 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.)
Ef = 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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