A body of mass `(5 pm 0.5)` kg is moving with a velocity of `(20pm0.4)`m/s . its kinetic energy will be

Options

A. `(1000 pm 0.14) J`
B. `(1000 pm 140) J`
C. `(500 pm 140) J`
D. `(500 pm 0.14) J`

Answer

To find the kinetic energy of a body, we use the formula:

\[ KE = \frac{1}{2} m v^2 \]

where \( m \) is the mass of the body and \( v \) is its velocity.

### Step 1: Substitute the values into the formula
Given:
- Mass \( m = 5 \pm 0.5 \) kg
- Velocity \( v = 20 \pm 0.4 \) m/s

Substituting the central values:
\[ KE = \frac{1}{2} \times 5 \times (20)^2 \]

### Step 2: Calculate \( v^2 \)
Calculating \( v^2 \):
\[ v^2 = 20^2 = 400 \]

### Step 3: Calculate \( KE \)
Now substituting \( v^2 \) back into the kinetic energy formula:
\[ KE = \frac{1}{2} \times 5 \times 400 \]
\[ KE = \frac{2000}{2} = 1000 \text{ Joules} \]

### Step 4: Calculate the uncertainty in kinetic energy
To find the uncertainty in kinetic energy \( \Delta KE \), we use the formula for error propagation:
\[ \frac{\Delta KE}{KE} = \frac{\Delta m}{m} + 2 \frac{\Delta v}{v} \]

Where:
- \( \Delta m = 0.5 \) kg
- \( m = 5 \) kg
- \( \Delta v = 0.4 \) m/s
- \( v = 20 \) m/s

### Step 5: Substitute the values into the error propagation formula
Calculating each term:
\[ \frac{\Delta m}{m} = \frac{0.5}{5} = 0.1 \]
\[ 2 \frac{\Delta v}{v} = 2 \times \frac{0.4}{20} = 2 \times 0.02 = 0.04 \]

Adding these contributions:
\[ \frac{\Delta KE}{KE} = 0.1 + 0.04 = 0.14 \]

### Step 6: Calculate the uncertainty in kinetic energy
Now, we can find \( \Delta KE \):
\[ \Delta KE = KE \times 0.14 = 1000 \times 0.14 = 140 \text{ Joules} \]

### Final Result
Thus, the kinetic energy of the body is:
\[ KE = 1000 \pm 140 \text{ Joules} \]