A body of mass 10 kg moves between two points on a horizontal road and does 500 J of work against friction. If the initial speed of the body is \( 20 \mathrm{~ms}^{-1} \), calculate its final speed

To find the final speed of the body, we can use the work-energy principle, which states that the work done on an object is equal to the change in its kinetic energy. 1. The initial kinetic energy (KE_initial) can be calculated with the formula: \[ KE_{\text{initial}} = \frac{1}{2} m v^2 = \frac{1}{2} \times 10 \, \text{kg} \times (20 \, \text{ms}^{-1})^2 = \frac{1}{2} \times 10 \times 400 = 2000 \, \text{J} \] 2. The work done against friction (W) is given as \(500 \, \text{J}\). Since this work is done against the motion, it will reduce the kinetic energy: \[ KE_{\text{final}} = KE_{\text{initial}} - W = 2000 \, \text{J} - 500 \, \text{J} = 1500 \, \text{J} \] 3. To find the final speed \(v_f\), we can use the kinetic energy formula: \[ KE_{\text{final}} = \frac{1}{2} m v_f^2 \implies 1500 \, \text{J} = \frac{1}{2} \times 10 \, \text{kg} \times v_f^2 \] 4. Solving for \(v_f^2\): \[ 1500 = 5 v_f^2 \implies v_f^2 = \frac{1500}{5} = 300 \] 5. Finally, taking the square root gives: \[ v_f = \sqrt{300} \approx 17.32 \, \text{ms}^{-1} \] Therefore, the final speed of the body is approximately \( 17.32 \, \text{ms}^{-1} \).