Question

a motor of input power 160 w raises a mass of 8.0 kg vertically at a constant speed of 0.50 m s⁻¹. what is the efficiency of the system?
a. 0.63 %
b. 25 %
c. 50 %
d. 100 %

Explanation:

Step1: Calculate the useful power output

The force required to lift the mass at a constant - speed is equal to its weight, $F = mg$, where $m = 8.0\ kg$ and $g= 10\ m/s^{2}$ (approximate value of gravitational acceleration). So $F=8.0\times10 = 80\ N$. The power formula is $P = Fv$. Given $v = 0.50\ m/s$, the useful power output $P_{out}=Fv=80\times0.50 = 40\ W$.

Step2: Calculate the efficiency

The efficiency $\eta$ of a system is given by the formula $\eta=\frac{P_{out}}{P_{in}}\times100\%$, where $P_{in}=160\ W$ and $P_{out} = 40\ W$. Substitute the values into the formula: $\eta=\frac{40}{160}\times100\% = 25\%$.

Answer:

B. 25%