A body of mass 5 kg has linear momentum of 20 kgm/s. If a constant force of 5N acts on it in the direction of its motion for 10s. The change in kinetic energy of the body is

Options

A. 320J
B. 600J
C. 200J
D. 450J

Answer

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

### Step 1: Find the Initial Velocity
The linear momentum \( p \) is given by the formula:
\[ p = m \cdot v \]
Where:
- \( p = 20 \, \text{kg m/s} \) (given)
- \( m = 5 \, \text{kg} \) (given)

We can rearrange the formula to find the initial velocity \( v \):
\[ v = \frac{p}{m} = \frac{20 \, \text{kg m/s}}{5 \, \text{kg}} = 4 \, \text{m/s} \]

### Step 2: Calculate the Acceleration
Using Newton's second law, the force \( F \) is related to mass \( m \) and acceleration \( a \) by:
\[ F = m \cdot a \]
Where:
- \( F = 5 \, \text{N} \) (given)

Rearranging to find acceleration:
\[ a = \frac{F}{m} = \frac{5 \, \text{N}}{5 \, \text{kg}} = 1 \, \text{m/s}^2 \]

### Step 3: Find the Final Velocity
We can use the first equation of motion to find the final velocity \( v_f \):
\[ v_f = v_i + a \cdot t \]
Where:
- \( v_i = 4 \, \text{m/s} \) (initial velocity from Step 1)
- \( a = 1 \, \text{m/s}^2 \) (acceleration from Step 2)
- \( t = 10 \, \text{s} \) (given)

Substituting the values:
\[ v_f = 4 \, \text{m/s} + (1 \, \text{m/s}^2 \cdot 10 \, \text{s}) = 4 \, \text{m/s} + 10 \, \text{m/s} = 14 \, \text{m/s} \]

### Step 4: Calculate the Change in Kinetic Energy
The change in kinetic energy \( \Delta KE \) can be calculated using the formula:
\[ \Delta KE = \frac{1}{2} m (v_f^2 - v_i^2) \]

Substituting the values:
- \( m = 5 \, \text{kg} \)
- \( v_f = 14 \, \text{m/s} \)
- \( v_i = 4 \, \text{m/s} \)

Calculating \( v_f^2 \) and \( v_i^2 \):
\[ v_f^2 = 14^2 = 196 \]
\[ v_i^2 = 4^2 = 16 \]

Now substituting into the change in kinetic energy formula:
\[ \Delta KE = \frac{1}{2} \cdot 5 \cdot (196 - 16) \]
\[ \Delta KE = \frac{1}{2} \cdot 5 \cdot 180 \]
\[ \Delta KE = \frac{5 \cdot 180}{2} = \frac{900}{2} = 450 \, \text{J} \]

### Final Answer
The change in kinetic energy of the body is:
\[ \Delta KE = 450 \, \text{J} \]

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