Physics Problem - 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 V₀=0 V=1m/s t=1s pₐ=?	K=½MV² K₀=0 pₐ=W/t W= K-K₀	pₐ=W/t=(K-K₀)/t=½MV²/t pₐ = ½MV²/t pₐ= 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) V₀=0 V=2m/s t=4s pₐ=? (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 or 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