#1 Stretching a certain spring 0.10 m from its relaxed length requires 30 J of w
ID: 1354142 • Letter: #
Question
#1
Stretching a certain spring 0.10 m from its relaxed length requires 30 J of work.
a) How much more work does it take to stretch this spring an additional 0.10 m?
#2
A motor must lift a 1300-kg elevator cab. The cab's maximum occupant capacity is 400 kg, and its constant "cruising" speed is 1.6 m/s . The design criterion is that the cab must achieve this speed within 2.0 s at constant acceleration, beginning from rest.
a) When the cab is carrying its maximum capacity, at what average rate must the motor deliver energy to get the cab up to cruising speed?
Explanation / Answer
1.
Energy required to stretch a spring, E = 1/2 kx2.
Where k is the spring constant, x is the distance to which it is stretched.
Given that, when x = 0.1 m, E = 30 J
1/2 k (0.1)2 = 30
k ( 0.01) = 60
k = 6000 N/m
In the second case, x = 0.1 + 0.1 = 0.2 m
E = 1/2 k x (0.2)2
= 1/2 x 6000 x 0.04 = 120 J
2.
Power = Force x velocity
Mass of elevator = 1300 kg
Mass of elevator + occupants, M = 1300 + 400 = 1700 kg.
Force, F = Mg = 1700 x 9.8 = 16660 N
Power = F x v
= 16660 x 1.6 = 26656 J/s.
Average power = 1/2 x 26656 = 13328 J/s
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