A hammer of mass 500 g, moving at 50 m s-1, strikes a nail. The nail stops the hammer in a very short time of 0.01 s. What is the force of the nail on the hammer

Asked by Vishal kumar | 2 years ago |  489

1 Answer

Solution :-

Given, mass of the hammer (m) = 500g = 0.5 kg

Initial velocity of the hammer (u) = 50 m/s

Terminal velocity of the hammer (v) = 0 (the hammer is stopped and reaches a position of rest).

Time period (t) = 0.01s

Therefore, the acceleration of the hammer is given by a \( = \frac{v-u}{t}= \frac{0-50m.s^{-1}}{0.01s}\)

a = -5000 ms-2

Therefore, the force exerted by the hammer on the nail (F = ma) can be calculated as:

F = (0.5 kg) * (-5000 ms-2) = -2500 N

As per the third law of motion, the nail exerts an equal and opposite force on the hammer. Since the force exerted on the nail by the hammer is -2500 N, the force exerted on the hammer by the nail will be +2500 N

Answered by Shivani Kumari | 2 years ago

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