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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?

Respuesta :

Answer:

a) 5908.02 W

b) 5250.855 W

Explanation:

[tex]v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{1.75-0}{3}\\\Rightarrow a=0.583\ m/s^2[/tex]

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow s=0\times t+\frac{1}{2}\times 0.583\times 3^2\\\Rightarrow s=2.6235\ m[/tex]

Change in Kinetic energy

[tex]\Delta KE=\frac{1}{2}m(v^2-u^2)\\\Rightarrow \Delta KE=\frac{1}{2}650(1.75^2-0^2)\\\Rightarrow \Delta=995.3125\ J[/tex]

Work

[tex]W=\Delta KE+U\\\Rightarrow W=995.3125+650\times 9.81\times 2.6235\\\Rightarrow W=17724.06025\ J[/tex]

Power

[tex]P=\frac{W}{t}\\\Rightarrow P=\frac{17724.06025}{3}\\\Rightarrow P=5908.02\ W[/tex]

Power is 5908.02 W

At cruising speed

[tex]P=\frac{W}{t}\\\Rightarrow P=\frac{F\times s}{t}\\\Rightarrow P=\frac{650\times 9.81\times 1.75\times 3}{3}\\\Rightarrow P=11158.875\ W[/tex]

The power is 11158.875 W

Difference in power is 11158.875-5908.02 = 5250.855 W