A machine, which is 72 percent efficient, uses 36 joules of energy in lifting up 1kg mass through a certain distance. The mass is the allowed to fall through that distance. The velocity at the end of its fall is

Options

A. `6.6 ms^(-1)`
B. `7.2 ms^(-1)`
C. `8.1 ms^(-1)`
D. `9.2 ms^(-1)`

Answer

To solve the problem step by step, we will follow these steps:

### Step 1: Understand the Work Done by the Machine
The machine is 72% efficient and uses 36 joules of energy to lift a 1 kg mass. The work done by the machine can be calculated using the efficiency formula:

\[ \text{Efficiency} = \frac{\text{Useful Work Output}}{\text{Total Work Input}} \times 100 \]

Given that the efficiency is 72%, we can rearrange the formula to find the useful work output (which is the potential energy gained by the mass):

\[ \text{Useful Work Output} = \text{Efficiency} \times \text{Total Work Input} \]

### Step 2: Calculate the Useful Work Output
Substituting the values into the equation:

\[ \text{Useful Work Output} = \frac{72}{100} \times 36 \text{ joules} \]

Calculating this gives:

\[ \text{Useful Work Output} = 0.72 \times 36 = 25.92 \text{ joules} \]

### Step 3: Relate Work Done to Potential Energy
The work done on the mass is equal to the change in potential energy when it is lifted:

\[ mgh = 25.92 \text{ joules} \]

Where:
- \( m = 1 \text{ kg} \)
- \( g = 9.81 \text{ m/s}^2 \) (acceleration due to gravity)
- \( h \) is the height lifted.

### Step 4: Solve for Height \( h \)
Rearranging the equation to find \( h \):

\[ h = \frac{25.92}{mg} = \frac{25.92}{1 \times 9.81} \approx 2.64 \text{ meters} \]

### Step 5: Calculate the Velocity at the End of the Fall
When the mass falls back down, the potential energy converts back into kinetic energy. The potential energy at the height \( h \) will equal the kinetic energy just before it hits the ground:

\[ mgh = \frac{1}{2} mv^2 \]

### Step 6: Cancel Mass and Solve for Velocity \( v \)
Since mass \( m \) appears on both sides, we can cancel it out:

\[ gh = \frac{1}{2} v^2 \]

Rearranging gives:

\[ v^2 = 2gh \]

Substituting the values:

\[ v^2 = 2 \times 9.81 \times 2.64 \]

Calculating this gives:

\[ v^2 = 51.8 \implies v \approx \sqrt{51.8} \approx 7.2 \text{ m/s} \]

### Final Answer
The velocity at the end of its fall is approximately **7.2 m/s**.

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