A block of mass 5 kg is raised from the bottom of the lake to a height of 3m without change in kinetic energy. If the density of the block is `3000 kg m^(-3)`, then the work done is equal to.

Options

A. `100 J`
B. `150 J`
C. `50 J`
D. `75 J`

Explanation

To solve the problem, we need to calculate the work done in raising a block of mass 5 kg from the bottom of a lake to a height of 3 meters, considering the density of the block is 3000 kg/m³.

### Step-by-Step Solution:

**Step 1: Calculate the volume of the block.**
The volume \( V \) of the block can be calculated using the formula:
\[ V = \frac{m}{\rho} \]
where \( m \) is the mass of the block and \( \rho \) is the density of the block.

Given:
- Mass \( m = 5 \, \text{kg} \)
- Density \( \rho = 3000 \, \text{kg/m}^3 \)

Substituting the values:
\[ V = \frac{5 \, \text{kg}}{3000 \, \text{kg/m}^3} = \frac{5}{3000} = \frac{1}{600} \, \text{m}^3 \]

**Step 2: Calculate the upthrust (buoyant force) acting on the block.**
The upthrust \( F_{\text{thrust}} \) can be calculated using Archimedes' principle:
\[ F_{\text{thrust}} = V \cdot \rho_{\text{water}} \cdot g \]
where \( \rho_{\text{water}} \) is the density of water (approximately \( 1000 \, \text{kg/m}^3 \)) and \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)).

Substituting the values:
\[ F_{\text{thrust}} = \left(\frac{1}{600} \, \text{m}^3\right) \cdot (1000 \, \text{kg/m}^3) \cdot (10 \, \text{m/s}^2) = \frac{10000}{600} = \frac{50}{3} \, \text{N} \]

**Step 3: Calculate the weight of the block.**
The weight \( W \) of the block is given by:
\[ W = m \cdot g \]
Substituting the values:
\[ W = 5 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 50 \, \text{N} \]

**Step 4: Calculate the net force acting on the block.**
The net force \( F \) acting on the block when it is raised can be calculated as:
\[ F = W - F_{\text{thrust}} \]
Substituting the values:
\[ F = 50 \, \text{N} - \frac{50}{3} \, \text{N} = \frac{150}{3} - \frac{50}{3} = \frac{100}{3} \, \text{N} \]

**Step 5: Calculate the work done in raising the block.**
The work done \( W_{\text{done}} \) is given by:
\[ W_{\text{done}} = F \cdot d \]
where \( d \) is the displacement (height raised, which is 3 m).

Substituting the values:
\[ W_{\text{done}} = \left(\frac{100}{3} \, \text{N}\right) \cdot (3 \, \text{m}) = 100 \, \text{J} \]

### Final Answer:
The work done in raising the block is **100 Joules**.

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