A bread gives a boy of mass 40 kg an energy of 21 kJ. If the efficiency is 28%, then the height that can be climbed by him using this energy, is :

Options

A. 22.5 m
B. 15 m
C. 10 m
D. 5 m

Answer

To solve the problem, we need to determine the height that a boy can climb using the energy obtained from a bread, given the efficiency of energy conversion. Here’s the step-by-step solution:

### Step 1: Calculate the Useful Energy
The total energy provided by the bread is given as 21 kJ. However, since the efficiency is only 28%, we need to calculate the useful energy that the boy can actually use.

\[ \text{Useful Energy} = \text{Total Energy} \times \text{Efficiency} \]

Converting 21 kJ to joules:
\[ 21 \text{ kJ} = 21,000 \text{ J} \]

Now, applying the efficiency:
\[ \text{Useful Energy} = 21,000 \text{ J} \times 0.28 = 5,880 \text{ J} \]

### Step 2: Relate Work Done to Potential Energy
When the boy climbs a height \( h \), the work done by him is equal to the increase in potential energy. The potential energy gained by climbing a height \( h \) is given by:

\[ \text{Potential Energy} = mgh \]

Where:
- \( m \) = mass of the boy = 40 kg
- \( g \) = acceleration due to gravity ≈ 9.8 m/s²
- \( h \) = height climbed

### Step 3: Set Up the Equation
Setting the useful energy equal to the potential energy gained:

\[ 5,880 \text{ J} = mgh \]

Substituting the values of \( m \) and \( g \):

\[ 5,880 \text{ J} = 40 \text{ kg} \times 9.8 \text{ m/s}^2 \times h \]

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

\[ h = \frac{5,880 \text{ J}}{40 \text{ kg} \times 9.8 \text{ m/s}^2} \]

Calculating the denominator:

\[ 40 \times 9.8 = 392 \text{ kg m/s}^2 \]

Now substituting back to find \( h \):

\[ h = \frac{5,880}{392} \approx 15 \text{ m} \]

### Conclusion
The height that the boy can climb using the energy from the bread is approximately **15 meters**.

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