Question

A lever requires an energy input of 10,000 J to move a 35 kg box on Earth (acceleration due to gravity =
9.81 m/s^2) up a distance of 18 m. What is the efficiency of this lever? Show all work

Answer 1

To determine the efficiency of the lever, we need to first calculate the useful work output and then use that to find the efficiency.

### Step 1: Calculate the Work Output

The work output (useful work done) when lifting the box can be calculated using the formula:

\[
W = F \cdot d
\]

where:
- \( W \) is the work done,
- \( F \) is the force applied,
- \( d \) is the distance moved.

#### Calculate the Force

The force required to lift the 35 kg box is equal to the weight of the box, which is given by:

\[
F = m \cdot g
\]

where:
- \( m = 35 \) kg (mass of the box),
- \( g = 9.81 \) m/s² (acceleration due to gravity).

Plugging in the values:

\[
F = 35 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 343.35 \, \text{N}
\]

#### Calculate the Work Done

Now, we can calculate the work done by lifting the box a distance of \( d = 18 \) m:

\[
W = 343.35 \, \text{N} \cdot 18 \, \text{m} = 6180.3 \, \text{J}
\]

### Step 2: Calculate the Efficiency

Efficiency (\( \eta \)) is given by the ratio of useful work output to the energy input, expressed as a percentage:

\[
\eta = \left( \frac{\text{Useful Work Output}}{\text{Energy Input}} \right) \times 100
\]

We know from the problem:
- Energy Input = 10,000 J,
- Useful Work Output = 6180.3 J (calculated above).

Now we can substitute these values into the efficiency formula:

\[
\eta = \left( \frac{6180.3 \, \text{J}}{10000 \, \text{J}} \right) \times 100
\]

\[
\eta = \left( 0.61803 \right) \times 100 = 61.803\%
\]

### Conclusion

The efficiency of the lever is:

\[
\eta \approx 61.8\%
\]