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Яндекс.Метрика

Thursday, September 29, 2016

Average Power

7.36. The electric motor of a 2-kg train accelerates the train from rest to 1 m/s in 1 s. Find the average power delivered to the train during the acceleration.

Solution

Given Data Useful formulas
M=2kg
V=0
V=1m/s
t=1s
p=?
K=½MV²
K₀=0
pₐ=W/t
W= K-K₀         
pₐ=W/t=(K-K₀)/t=½MV²/t
pₐ = ½MV²/t
p= ½ (2kg) (1m/s)² /1s =
=1 kg m/s²∙m/s=1N∙m/s=1J/s=1W
pₐ=1W

7.40. A 1000-kg elevator starts from rest. It moves upward for 4 s with constant acceleration until it reaches its cruising speed of 2 m/s.
(a) What is the average power of the elevator motor during this period?
(b) What is the motor power when the elevator moves at its cruising speed?

Given Data Useful formulas
M=1000kg
(a)
V=0
V=2m/s
t=4s
p=?



(b)
V=const=2m/s
K=½MV²
K₀=0
U=Mgh
pₐ=W/t
W= (K+U)-(K₀+ U₀)
h=V₀t+at²/2
a=(V-V₀)/t



p = F V = MgV
W= (K+U)-(K₀+ U₀)= K+U=½MV²+Mgh=½MV²+Mgh=
h=at²/2=(V/t)∙t²/2=Vt/2
W= ½MV²+MgVt/2
pₐ=W/t=½MV²/t+MgV/2
pₐ=W/t=M(V/t+g)½V
½MV²/t=½(1000kg)(2m/s)²/4s=500W
MgV/2=1000kg∙10m/s²∙2m/s /2=10000W
pₐ=10500W=10.5kW
or pₐ=1000kg ∙ (2 m/s / 4 s + 10 m/s²) ½ ∙ 2 m/s = 10.5kW

p=MgV=1000kg∙10m/s²∙2m/s=20000W=20kW





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