Question

1.A 650-kg elevator starts from rest. It moves upward for3 s with constant acceleration until it reaches its cruising speed, 1.75 m/s. (a) What is the average power of the elevator motor during this period? (b) How does this compare with its power during an upward cruise with constant speed?

Answer 1

a) 5908.02 W

b) 5250.855 W

Step-by-step explanation:

`v=u+at\\\Rightarrow a=(v-u)/(t)\\\Rightarrow a=(1.75-0)/(3)\\\Rightarrow a=0.583\ m/s^2`

`s=ut+(1)/(2)at^2\\\Rightarrow s=0* t+(1)/(2)* 0.583* 3^2\\\Rightarrow s=2.6235\ m`

Change in Kinetic energy

`\Delta KE=(1)/(2)m(v^2-u^2)\\\Rightarrow \Delta KE=(1)/(2)650(1.75^2-0^2)\\\Rightarrow \Delta=995.3125\ J`

Work

`W=\Delta KE+U\\\Rightarrow W=995.3125+650* 9.81* 2.6235\\\Rightarrow W=17724.06025\ J`

Power

`P=(W)/(t)\\\Rightarrow P=(17724.06025)/(3)\\\Rightarrow P=5908.02\ W`

Power is 5908.02 W

At cruising speed

`P=(W)/(t)\\\Rightarrow P=(F* s)/(t)\\\Rightarrow P=(650* 9.81* 1.75* 3)/(3)\\\Rightarrow P=11158.875\ W`

The power is 11158.875 W

Difference in power is 11158.875-5908.02 = 5250.855 W