The head of a 3.2 kg hammer, moving at 5.2 m/s, strikes a nail and drives it int

Question

The head of a 3.2 kg hammer, moving at 5.2 m/s, strikes a nail and drives it into hardwood. (a.) The head stays in contact for 2.2 ms and rebounds with negligible velocity. What is the average force exerted on the nail?  (b). What is the change in kinetic energy of the hammer? (c.) What was the impulse exerted on the nail by the hammer? (d) When the same hammer hits a springy nail, it rebounds with its initial speed. The contact time is the same. What force is exerted this time?  (e) What is the change in kinetic energy of the hammer?  (f) What is the impulse?

Explanation / Answer

mass m = 3.2 kg initial speed u = 5.2 m / s final speed v = 0 time of contact t = 2.2 * 10 ^ -3 s Average force F = m ( u - v ) / t                          = 7.5636 * 10 ^ 3 N (b). Change in kinetic energy of the hammer = ( 1/ 2) m[ u ^ 2-v ^ 2]                                                                     =43.264 J (c). Impulse J = m ( u - v )                      = 16.64 kg m / s (d). final speed v = -5.2 m / s force F = m ( v - u ) / t            =15.12 * 10 ^ 3 N (e). change in kinetic energy = ( 1/ 2) m [ v ^ 2 - u ^ 2]                                           = 0   Since magnitude of u and v are same (f). impulse J = m ( u - v ) = 33.28 kg m / s

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