contestada

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?

Respuesta :

Part a)

Power rated on the elevator is given as

[tex]P = 8 kW[/tex]

[tex]P = 8 \times 10^3 W[/tex]

now the mass that is lifted above is given as

[tex]m = 1225 kg[/tex]

height of the elevator lifted is

[tex]h = 9 m[/tex]

now the potential energy is given as

[tex]U = mgh[/tex]

[tex]U = 1225 (9.8)(9) = 108045 J[/tex]

now power is defined as rate of energy

[tex]P = \frac{W}{t}[/tex]

[tex]8 \times 10^3 = \frac{108045}{t}[/tex]

[tex]t = 13.5 s[/tex]

so it will take 13.5 s to lift up

Part b)

Electrical energy used

[tex]efficiency = 90[/tex]

[tex]0.90 = \frac{Output}{Input}[/tex]

[tex]0.90 = \frac{108045}{Input}[/tex]

[tex]Input = 120050 J[/tex]

so electrical energy used in this process will be 120050 J

Otras preguntas