.An elevator of mass 1000 kg accelerates from rest moving upward. During the fir
ID: 1792270 • Letter: #
Question
.An elevator of mass 1000 kg accelerates from rest moving upward. During the first few seconds of motion, the elevator's acceleration increases with time according to the expressiol a 2.00t-0.300f where t is in seconds and a is in m/s2 (a) What is the change in kinetic energy of the elevator during the interval from 0 to t = 2.00 s? [5 pts] (b) How far does the elevator rise in this time? [4 pts] (c) What is the minimum average power output of the motor lifting the elevator over this time interval? [3 pts] (d) Why might the average power actually supplied be larger that the minimum value calculated in the previous part? [1 pt]Explanation / Answer
a)
a = 2t - 0.3 t2
dv/dt = 2t - 0.3 t2
dv = (2t - 0.3 t2 ) dt
integrating both side
vf - vo = t2 - (0.1) t3 = (2)2 - (0.1) (2)3 = 3.2
vo = 0
hence vf = 3.2 m/s
change in KE = (0.5) m (vf2 - vo2) = (0.5) (1000) (3.22 - 02) = 5120 J
b)
v = t2 - (0.1) t3
dx/dt = t2 - (0.1) t3
taking derivative both side
x = t3/3 - (0.025) t4
x = (2)3/3 - (0.025) (2)4
x = 2.3 m
c)
Pavg = change in KE /t = 5120/2 = 2560 Watt
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