Question

A body of mass \( 5 \, \text{kg} \) is moving with a momentum of \( 10 \, \text{kg} \cdot \text{ms}^{-1} \). Now a force of \( 2 \, \text{N} \) acts on the body in the direction of its motion for \( 5 \, \text{s} \). The increase in the kinetic energy of the body is _____ J.

Answer 1

The kinetic energy of a body is given by: \[ KE = \frac{p^2}{2m}, \] where: 
\( p \) is the momentum, 
\( m = 5 \, \text{kg} \) is the mass of the body. 
Step 1: Initial kinetic energy. The initial momentum is \( p_1 = 10 \, \text{kg} \cdot \text{ms}^{-1} \). The initial kinetic energy is: \[ KE_1 = \frac{p_1^2}{2m} = \frac{(10)^2}{2 \cdot 5} = \frac{100}{10} = 10 \, \text{J}. \] 
Step 2: Final kinetic energy. The force \( F = 2 \, \text{N} \) acts for \( t = 5 \, \text{s} \), producing an additional momentum: \[ \Delta p = F \cdot t = 2 \cdot 5 = 10 \, \text{kg} \cdot \text{ms}^{-1}. \] The final momentum is: \[ p_2 = p_1 + \Delta p = 10 + 10 = 20 \, \text{kg} \cdot \text{ms}^{-1}. \] The final kinetic energy is: \[ KE_2 = \frac{p_2^2}{2m} = \frac{(20)^2}{2 \cdot 5} = \frac{400}{10} = 40 \, \text{J}. \] 
Step 3: Increase in kinetic energy. The increase in kinetic energy is: \[ \Delta KE = KE_2 - KE_1 = 40 - 10 = 30 \, \text{J}. \] 
Final Answer: The increase in kinetic energy is: \[ \boxed{30 \, \text{J}}. \]