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3. A 1000-kg elevator is moving from `10\ m` above the ground to `20\ m` above the ground.
(a) What is the initial potential energy of the elevator.
(b) What is the final potential energy of the elevator.
(c) Calculate the work done by the lifting force.
3. A 1000-kg elevator is moving from 10\ m above the ground to 20\ m above the ground.
(a) What is the initial potential energy of the elevator.
(b) What is the final potential energy of the elevator.
(c) Calculate the work done by the lifting force. `PE = mgh`
`W = ∆PE`
(a)
`PE_i = 1.0 times 10^5\ J`
(b)
`PE_f = 2.0 times 10^5\ J`
(c)
`W = 1.0 times 10^5\ J`
Given
`m = 1000\ kg`
`h_i = 10\ m`
`h_f = 20\ m`
Equations:
`PE = mgh`
`W = ∆PE`
Solution
(a)
`PE_i = mgh_i`
`PE_i = (1000\ kg)(10\ m//s^2)(10\ m)`
`PE_i = 1.0 times 10^5\ J`
(b)
`PE_f = mgh_f`
`PE_f = (1000\ kg)(10\ m//s^2)(20\ m)`
`PE_f = 2.0 times 10^5\ J`
(c)
`W = ∆PE = PE_f - PE_i`
`W = ( 2.0 times 10^5\ J)-( 1.0 times 10^5\ J)`
`W = 1.0 times 10^5\ J` |