An elevator (mass M = 2720 kg) is moving downward at v0 = 2.52 m/s.A braking sys

Question

An elevator (mass M = 2720 kg) is moving downward at v0 = 2.52 m/s.A braking system prevents the downward speed from increasing. Theacceleration of gravity is 9.8 m/s2 .
At what rate is the braking system converting mechanical energy tothermal energy?
Answer in units of kW.

While the elevator is moving downward at v0 = 2.52 m/s, the brakingsystem fails and the elevator is in free-fall for a distance d =1.22 m before hitting the top of a large safety spring with forceconstant
k = 16008 N/m. After the elevator hits the top of the spring, wewant to know the dis-
tance y that the spring is compressed before the elevator isbrought to rest. Writ an alge-
braic expression for the value of y in terms of the knownquantities M, vo, g, k, and d,
and substitute the given values to find y.

Answer in units of m.

Explanation / Answer

a)P=FV0 F=Mg P=Mg V0 P= 2720 x 9.8 x 2.52 P=67.2 kW b) Pe= Ps Pe=Mgh Ps= 0.5k y2 Mgh=0.5k y2 y = [2Mgh/k] y = [2 x 2720 x 9.8 x1.22/16008] y = 2.02 m y = [2 x 2720 x 9.8 x1.22/16008] y = 2.02 m

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