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Here is a quick guide to doing problems involving power , to help you do your physics homework or assignment:
The average power delivered to a system is equal to the amount of energy transfer over a period time,
Examples:
1. The electric motor of a model train accelerates the train from rest to 0.620 m/s in 21.0 ms. The total mass of the train is 875 g. Find the average power delivered to the train during its acceleration.
Solution
The average power delivered to the train is its change in kinetic energy over the time it takes,
(0.620 \, m/s)^2-\frac{1}{2}(0.875\,kg)(0 \, m/s)^2}{0.021 \, s }=8 \, W "P=\frac{W}{t}=\frac{\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2}{t}=\frac{\frac{1}{2}(0.875 \, kg)(0.620 \, m/s)^2-\frac{1}{2}(0.875\,kg)(0 \, m/s)^2}{0.021 \, s }=8 \, W")
2. A skier of mass 70 kg is pulled up a slope by a motor driven cable. (a) How much work is required to pull him 60 m up a 30o slope (assumed frictionless) at a constant speed of 2.0 m/s? (b) What power must a motor have to perform this task?
Solution
a) Work done by the is the force in the cable multiplied by the distance that it pulls the skier.
The tension in the cable is:
and the work done is
(9.8\, m/s^2)(60\,m)sin\,30^o=2.1 \times 10^4 \, J "W=Td = mgd sin \theta = (70 \, kg)(9.8\, m/s^2)(60\,m)sin\,30^o=2.1 \times 10^4 \, J")
Hence the work required to pull the skier up the slope is 21000 Joules
b)
The power of the motor to pull the skier is,
(9.8\,m/s^2)(sin\,30^o)(2.0\,m/s)=686 W "P=(70\,kg)(9.8\,m/s^2)(sin\,30^o)(2.0\,m/s)=686 W")
Hence the average power given by the motor is 686 Watts.
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