An elevator's electric hoist is rated at 8.0 kw and has an efficiency of 90.0%. the hoist lifts the elevator car, mass 1225 kg, a distance of 9.00 m. (a) how much time is required? (b) how much electrical energy is used to perform this task?

Part a)

Power rated on the elevator is given as

now the mass that is lifted above is given as

height of the elevator lifted is

now the potential energy is given as

now power is defined as rate of energy

so it will take 13.5 s to lift up

Part b)

Electrical energy used

so electrical energy used in this process will be 120050 J

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.