The hammer throw is a track and field event in which a 6.64-kg ball (the \"hamme
ID: 1978305 • Letter: T
Question
The hammer throw is a track and field event in which a 6.64-kg ball (the "hammer"), starting from rest, is whirled around in a circle several times and released. It then moves upward on the familiar curving path of projectile motion. In one throw, the hammer is given a speed of 33.5 m/s. For comparison, a .22 caliber bullet has a mass of about 2.78 g and, starting from rest, exits the barrel of a gun with a speed of 430 m/s. Determine the work done to launch the motion of (a) the hammer and (b) the bullet.Explanation / Answer
Given that Mass of hammer is M = 6.64 kg Speed of the hammer is u = 33.5 m/s Mass of the bullet is m = 2.78 g = ( 2.78 g ) ( 0.001 kg / 1 g ) Speed of the bullet is v = 430 m/s ------------------------------------------------------- Intially, the speed of the hammer and bullet is zero. According to work energy theorem, Work done = Change in kinetic energy a) The work done by the hammer is W = ( 1 /2 ) M u 2 = ( 0.5 ) ( 6.64 kg ) ( 33.5 m/s ) 2 = 3725.87 J ___________________________________________ b) The work done by the bullet is W' = (1 / 2 ) m v2 = (1 / 2) ( 2.780 * 10-3 kg) (430 m/s)2 = 257.011 J Given that Mass of hammer is M = 6.64 kg Speed of the hammer is u = 33.5 m/s Mass of the bullet is m = 2.78 g = ( 2.78 g ) ( 0.001 kg / 1 g ) Speed of the bullet is v = 430 m/s ------------------------------------------------------- Intially, the speed of the hammer and bullet is zero. According to work energy theorem, Work done = Change in kinetic energy a) The work done by the hammer is W = ( 1 /2 ) M u 2 = ( 0.5 ) ( 6.64 kg ) ( 33.5 m/s ) 2 = 3725.87 J ___________________________________________ b) The work done by the bullet is W' = (1 / 2 ) m v2 = (1 / 2) ( 2.780 * 10-3 kg) (430 m/s)2 = 257.011 JRelated Questions
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