A shot putter launches a 7.57 kg shot by pushing it along a straight line of len

Question

A shot putter launches a 7.57 kg shot by pushing it along a straight line of length 1.41 m and at an angle of 36.0° from the horizontal, accelerating the shot to the launch speed from its initial speed of 2.7 m/s (which is due to the athlete's preliminary motion). The shot leaves the hand at a height of 2.13 m and at an angle of 36.0°, and it lands at a horizontal distance of 15.6 m.What is the magnitude of the athlete's average force on the shot during the acceleration phase? (Hint: Treat the motion during the acceleration phase as though it were along a ramp at the given angle.)

Explanation / Answer

The mass of the short putter m = 7.57kg the initial veclotiy vi = 2.7m/s The horizontal distance X = 15.6m then 15.6 = v0 cos36 t          t = 15.6 /v0 cos36 and        2.13 = v0 sin36 t - 1/2 gt^2         2.13 = 15.6 tan36 - 0.5(9.8)(15.6 /v0 cos36)^2          2.13 = 11.33 - 1821.9/ v0 ^2           1821.9 = 9.2v0^2              v0 = 14m/s         From Newton's law     Fnet = F - mgsin 36 Now from work energy theorem       (F - mgsin36) *L = 1/2 m(vf^2 - vi^2)       F = mgsin36 + [(0.5)m(vf^2 - vi^2)]/L          = (7.57)(9.8)sin36 + (0.5)(7.57)[(14)^2 - (2.7)^2] /1.41         = 43.6 + 506.5 =   550N     

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