A force of 10 N acts on a body of mass 1 kg at rest for 10 s and the body acquire a kinetic energy `K_(1)` . A force of 10 N acts on a body of mass 1 kg at rest through a distance of 10 m and the body acquires a kinetic energy `K_(2)`. The ratio of `K_(1)` to `K_(2)` is

Options

A. `1:1`
B. `2:1`
C. `20:1`
D. `50:1`

Answer

To solve the problem, we need to find the kinetic energies \( K_1 \) and \( K_2 \) for two different scenarios and then calculate the ratio \( \frac{K_1}{K_2} \).

### Step 1: Calculate \( K_1 \)

1. **Identify the given values**:
- Force \( F = 10 \, \text{N} \)
- Mass \( m = 1 \, \text{kg} \)
- Time \( t = 10 \, \text{s} \)

2. **Calculate acceleration using Newton's second law**:
\[ F = m \cdot a \implies a = \frac{F}{m} = \frac{10 \, \text{N}}{1 \, \text{kg}} = 10 \, \text{m/s}^2 \]

3. **Use the first equation of motion to find final velocity \( v \)**:
\[ v = u + a \cdot t \]
Since the initial velocity \( u = 0 \):
\[ v = 0 + 10 \cdot 10 = 100 \, \text{m/s} \]

4. **Calculate kinetic energy \( K_1 \)**:
\[ K_1 = \frac{1}{2} m v^2 = \frac{1}{2} \cdot 1 \cdot (100)^2 = \frac{1}{2} \cdot 1 \cdot 10000 = 5000 \, \text{J} \]

### Step 2: Calculate \( K_2 \)

1. **Identify the given values for the second scenario**:
- Force \( F = 10 \, \text{N} \)
- Mass \( m = 1 \, \text{kg} \)
- Distance \( s = 10 \, \text{m} \)

2. **Use the same acceleration calculated earlier**:
\[ a = 10 \, \text{m/s}^2 \]

3. **Use the third equation of motion to find final velocity \( v \)**:
\[ v^2 = u^2 + 2as \]
Since the initial velocity \( u = 0 \):
\[ v^2 = 0 + 2 \cdot 10 \cdot 10 = 200 \]

4. **Calculate kinetic energy \( K_2 \)**:
\[ K_2 = \frac{1}{2} m v^2 = \frac{1}{2} \cdot 1 \cdot 200 = 100 \, \text{J} \]

### Step 3: Calculate the ratio \( \frac{K_1}{K_2} \)

1. **Find the ratio**:
\[ \frac{K_1}{K_2} = \frac{5000}{100} = 50 \]

Thus, the ratio of \( K_1 \) to \( K_2 \) is \( 50:1 \).

### Final Answer:
The ratio \( \frac{K_1}{K_2} = 50:1 \).

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