Question -
Answer -
Given, mass of thehammer (m) = 500g = 0.5kg
Initial velocity of thehammer (u) = 50 m/s
Terminal velocity ofthe hammer (v) = 0 (the hammer is stopped and reaches a position of rest).
Time period (t) =0.01s
a = -5000ms-2
Therefore, the forceexerted by the hammer on the nail (F = ma) can be calculated as:
F = (0.5kg) * (-5000ms-2) = -2500 N
As per the third lawof motion, the nail exerts an equal and opposite force on the hammer. Since theforce exerted on the nail by the hammer is -2500 N, the force exerted on thehammer by the nail will be +2500 N.