Question

The momentum of a body of mass 2 kg is 10 kg m/s. A force 2 N acts on the body in the direction of motion for 5 s. The increase in kinetic energy is:

• A. \( 100 \, {J} \)
• B. \( 75 \, {J} \)
• C. \( 125 \, {J} \)
• D. \( 50 \, {J} \)

Hint:

When a force acts on a body and causes a change in velocity, use the work-energy theorem to find the increase in kinetic energy.

Answer 1

The Correct Option is C

The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy.

Given:
- Mass of the body: \( m = 2 \, \text{kg} \)
- Initial momentum: \( p_{\text{initial}} = 2 \times 10 = 20 \, \text{kg m/s} \)
- Force applied: \( F = 2 \, \text{N} \)
- Time: \( t = 5 \, \text{s} \)
- Initial velocity: \( v_{\text{initial}} = 10 \, \text{m/s} \)

Step 1: Find acceleration
The acceleration \( a \) is:
\[ a = \frac{F}{m} = \frac{2}{2} = 1 \, \text{m/s}^2 \]

Step 2: Find final velocity
Using the equation \( v_{\text{final}} = v_{\text{initial}} + a \cdot t \), we get:
\[ v_{\text{final}} = 10 + 1 \times 5 = 15 \, \text{m/s} \]

Step 3: Calculate the increase in kinetic energy
The increase in kinetic energy is:
\[ \Delta KE = \frac{1}{2} m \left( v_{\text{final}}^2 - v_{\text{initial}}^2 \right) \] \[ \Delta KE = \frac{1}{2} \times 2 \left( 15^2 - 10^2 \right) \] \[ \Delta KE = 1 \times \left( 225 - 100 \right) = 125 \, \text{J} \]