Question

The operating efficiency of a 0.5 A, 120 V electric motor that lifts a 9 kg mass against gravity at an average
velocity of 0.5 m/s is most nearly
(A) 13% (B) 25% (C) 53% (D) 75 %

Answer 1

$\underline{ \boxed{ \bold{ \mathfrak{ \huge{ \purple{Answer}}}}}}$

Given :

Current in motor = 0.5 A

Volatage drop = 120 V

mass lifted by motor = 9 kg

average velocity = 0.5 m/s

To Find :

efficiency of motor...

Formula :

efficiency of motor is given by

$\:\:\:\:\dag \: \underline{ \boxed{ \bold{ \rm{ \pink{ \eta = \frac{output \: power}{input \: power } \times 100}}}}} \: \dag$

Formula of electrical power is given by

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \dag \: \underline{ \boxed{ \bold{ \rm{ \purple{P = VI}}}}} \: \dag$

Formula of mechanical power is given by

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \dag \: \underline{ \boxed{ \bold{ \rm{ \green{P = Fv}}}}} \: \dag$

Calculation :

$\leadsto \rm \: \eta = \frac{mechanical \: power}{electrical \: power} \times 100 = \frac{mg \times v}{VI} \times 100\\ \\ \leadsto \rm \: \eta = \frac{ 9 \times 10 \times 0.5}{120 \times 0.5} \times 100 = \frac{900}{12} \\ \\ \therefore \: \underline{ \boxed{ \bold{ \rm{ \orange{ \eta = 75\%}}}}} \: \red{ \star}$

Answer 2

Answer:

Given:

Current = 0.5 A

Voltage = 120 V

Mass lifted = 9 kg

Average Velocity = 0.5 m/s

To find:

Operating efficiency

Concept:

No machine have 100 % efficiency. This is because some energy is dissipated from the engine as heat and lost.

So, amount of the supplied energy is always greater than that of work done by the machine.

Calculation:

Supplied Power is the electrical Power given to the machine. Let it be denoted as SP.

$SP = current \times voltage \\ = > SP = 0.5 \times 120 \\ = > SP = 60 \: watt$

Mechanical power is the work done by machine per unit time. Let it be denoted by MP.

$MP = force \times velocity \\ = >MP = (m \times g) \times v \\ = > MP = 9 \times 10 \times (0.5) \\ = > MP = 45 \: watt$

Let efficiency be denoted as η.

$\eta = \frac{MP}{SP} \times 100\% \\ = > \eta = \frac{45}{60} \times 100\% \\ = > \eta = 75\%$

So final answer :

$\boxed{ \sf{ \green{ \bold{ \huge{ \eta = 75\%}}}}}$