Question

A force of $250\text{ N}$ is required to lift a mass of $75\text{ kg}$ through a pulley system. In order to lift this mass through $3\text{ m}$, the rope has to be pulled through $12\text{ m}$. The efficiency of the system is:

• A. $50\%$
• B. $75\%$
• C. $33\%$
• D. $90\%$

Hint:

Efficiency can also be written as:
$\eta = \frac{\text{Mechanical Advantage (MA)}}{\text{Velocity Ratio (VR)}} \times 100\%$.
Here, $\text{MA} = \frac{mg}{F} = \frac{750}{250} = 3$, and $\text{VR} = \frac{d}{h} = \frac{12}{3} = 4$.
Thus, $\eta = \frac{3}{4} \times 100\% = 75\%$.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to calculate the efficiency of a pulley system given the input force, input distance (rope pulled), load mass, and output distance (height raised).

Step 2: Key Formula and Approach:
Efficiency ($\eta$) of a machine is the ratio of useful work output to total work input, expressed as a percentage:
\[ \eta = \frac{W_{\text{out}}}{W_{\text{in}}} \times 100\% \] where $W_{\text{out}} = m g h$ and $W_{\text{in}} = F \times d$.

Step 3: Detailed Explanation:

• Calculate Work Input ($W_{\text{in}}$):
The force applied $F = 250\text{ N}$ acts through a distance $d = 12\text{ m}$.
\[ W_{\text{in}} = F \times d = 250\text{ N} \times 12\text{ m} = 3000\text{ J} \]

• Calculate Work Output ($W_{\text{out}}$):
The mass lifted $m = 75\text{ kg}$ is raised through a height $h = 3\text{ m}$.
Assuming $g = 10\text{ ms}^{-2}$:
\[ W_{\text{out}} = m g h = 75\text{ kg} \times 10\text{ ms}^{-2} \times 3\text{ m} = 2250\text{ J} \]

• Calculate Efficiency ($\eta$):
\[ \eta = \frac{2250}{3000} \times 100\% = 0.75 \times 100\% = 75\% \]

Step 4: Final Answer:
The efficiency of the pulley system is $75\%$, which corresponds to Option (B).