A student measures motor efficiency. The motor lifts a 2.0 N weight 0.80 m. Supply: 4.5 V, 0.30 A. Time: 6.0 s. (a) Calculate the useful work done. [2] (b) ...
Question 1 Report
A student measures motor efficiency. The motor lifts a 2.0 N weight 0.80 m. Supply: 4.5 V, 0.30 A. Time: 6.0 s.
(a) Calculate the useful work done. [2]
(b) Calculate the electrical energy supplied. [2]
(c) Calculate the efficiency. [2]
(d) With a 5.0 N weight taking 12 s, calculate the new efficiency. [2]
(e) State what happens to the wasted energy. [1]
Answer Details
(a) Useful work done [2]
The useful work is the gravitational potential energy gained by the weight as it is lifted:
\(W = F \times d = 2.0 \times 0.80\) [1]
\(W = 1.6\text{ J}\) [1]
(Here \(F\) is the weight in newtons and \(d\) is the height lifted.)
(b) Electrical energy supplied [2]
\(E = VIt = 4.5 \times 0.30 \times 6.0\) [1]
\(E = 8.1\text{ J}\) [1]
(c) Efficiency [2]
\(\text{Efficiency} = \dfrac{\text{useful output}}{\text{total input}} = \dfrac{1.6}{8.1}\) [1]
\(\text{Efficiency} = 0.198 \text{ or } 19.8\%\) [1]
About 80% of the electrical energy is wasted rather than doing useful work.
(d) New efficiency with a 5.0 N weight taking 12 s [2]
Useful work: \(W = 5.0 \times 0.80 = 4.0\text{ J}\)
Electrical input: \(E = 4.5 \times 0.30 \times 12 = 16.2\text{ J}\)
\(\text{Efficiency} = \dfrac{4.0}{16.2}\) [1]
\(\text{Efficiency} = 0.247 \text{ or } 24.7\%\) [1]
The efficiency has increased from 19.8% to 24.7% with the heavier load, because the motor is doing proportionally more useful work relative to its fixed overhead losses.
(e) What happens to the wasted energy [1]
The wasted energy is transferred to thermal energy in the motor windings and bearings (the motor gets warm), with a small amount also dissipated as sound energy. [1]
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