Question

A synchronous motor is used to lift an elevator and its load of 1500 kg to a height of 20 m. The time taken or job is 20 seconds .What is work done? Is the efficiency of the motor is 75% at which rate is the energy supplied to the motor ?

Answer 1

Hi

Here is your answer,

Given mass, m = 1500 kg, h = 20 m ,  η = 75% , t = 20 s 

Work done,            W = mgh = 1500 × 9.8 × 20
           
                                   = 2.94 × 10⁵ J 

 Rate of doing work = W/t = 2.94 × 10⁵/20

                 = 1.47 × 10⁴ W 

As, efficiency   η = Output power/Input power 

                 75/100 = 1.47 × 10⁴ / Input power

 Input power or the rate at which energy is supplied = 1.47 × 10⁴ × 100/ 75

 
                                     = 1.96 × 10⁴ W



Hope it helps you !

Answer 2

(i) 2.94×10⁵J

(ii) 1.96×10⁴W

Explanation:

Given,

m = 1500kg

h = 20m

t = 20s

n = 75% = 75/100

(i) W = mgh

= 1500×9.8×20

= 294000J

= 2.94×10⁵J

(ii) n = Output Power/Input power

Input Power = Output Power/n

P = W/t (Output Power)

P = 2.94×10⁵J/20s

= 1.47×10⁴W

Input Power = 1.47×10⁴/(75/100)

= 1.47×10⁴×100/75

= 1.96×10⁴W

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