Question

A weight lifting machine is used to lift a load of 70 kg through 30 cm distance. If 30 kg load is applied to move a distance 100 cm, find efficiency of the machine.

• A. 66%
• B. 70%
• C. 75%
• D. 80%

Hint:

Efficiency is a dimensionless ratio, so as long as the units for force (Load and Effort) are the same and the units for distance (Distance Load Moved and Distance Effort Moved) are the same, the units will cancel out. You don't necessarily have to convert to standard SI units (Newtons and meters), but it is good practice to avoid errors.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This question is about calculating the efficiency of a simple machine. The efficiency (\(\eta\)) of a machine is the ratio of the useful work output to the total work input. Work is calculated as Force \(\times\) Distance.

Step 2: Key Formula or Approach:
The formula for efficiency is:
\[ \eta = \frac{\text{Work Output}}{\text{Work Input}} \times 100\% \] Where:

• Work Output is the work done on the load. \( \text{Work Output} = \text{Load} \times \text{Distance Load Moved} \).
• Work Input is the work done by the effort. \( \text{Work Input} = \text{Effort} \times \text{Distance Effort Moved} \). In this problem, 'kg' is used as a unit of force (kg-force or kgf). We must also ensure the distance units are consistent.

Step 3: Detailed Explanation:
First, identify the given values:
For the Output:

• Load (L) = 70 kgf
• Distance Load Moved (\(d_L\)) = 30 cm = 0.3 m For the Input:

• Effort (E) = 30 kgf
• Distance Effort Moved (\(d_E\)) = 100 cm = 1.0 m Now, calculate the Work Output and Work Input. It's best to convert distances to meters to work in standard units (Joules), although the ratio will be the same if we keep them in cm. Let's use meters.
Calculate Work Output:
\[ \text{Work Output} = L \times d_L = 70 \text{ kgf} \times 0.3 \text{ m} = 21 \text{ kgf-m} \] (Note: kgf-m is a unit of work)
Calculate Work Input:
\[ \text{Work Input} = E \times d_E = 30 \text{ kgf} \times 1.0 \text{ m} = 30 \text{ kgf-m} \] Calculate Efficiency (\(\eta\)):
\[ \eta = \frac{\text{Work Output}}{\text{Work Input}} \times 100\% \] \[ \eta = \frac{21 \text{ kgf-m}}{30 \text{ kgf-m}} \times 100\% \] \[ \eta = \frac{21}{30} \times 100\% \] \[ \eta = \frac{7}{10} \times 100\% \] \[ \eta = 0.7 \times 100\% = 70\% \]

Step 4: Final Answer:
The efficiency of the machine is 70%.