Question

question 20 (1 point)

an elevator and its occupants have a mass of $1.25 \times 10^3$ kg. the elevator motor can lift this load a distance of 42.0 m in 15.0 s. if the motor uses $2.30 \times 10^6$ j of energy to do this, what is the efficiency of the system?

\\(\circ\\) 0.312
\\(\circ\\) 0.224
\\(\circ\\) 0.286
\\(\circ\\) 0.199

Explanation:

Step1: Calculate useful work done

Use work formula $W = mgh$.
$m=1.25\times10^3\ \text{kg}$, $g=9.8\ \text{m/s}^2$, $h=42.0\ \text{m}$
$W = 1.25\times10^3 \times 9.8 \times 42.0 = 5.145\times10^5\ \text{J}$

Step2: Compute system efficiency

Efficiency $\eta = \frac{\text{Useful Work}}{\text{Total Energy Input}}$
$\eta = \frac{5.145\times10^5}{2.30\times10^6}$

Step3: Simplify the ratio

$\eta = \frac{5.145}{23.0} \approx 0.224$

Answer:

0.224