A shot putter launches a 7.260kg shot by pushing it along a straight line of len

Question

A shot putter launches a 7.260kg shot by pushing it along a straight line of length 1.650 m and at an angle of 34.10 degree from the horizontal, accelerating the shot to the launch speed from its initial speed of 2.500m/s (which is due to the athlete's preliminary motion). The shot leaves the hand at a height of 2.110 m and at an angle of 34.10 degree, and it lands at horizontal distance of 15.90 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 given angle.)

Explanation / Answer

The net force on the shot is F-6.85*9.8sin34.3 where F is the force of the athlete and subtracting from that is the component of gravity in line with the athlete's force.

The horizontal distance traveled = Vxt = Vocos34.3*t

Where t is the time of flight which is twice the time to peak altitude Vyo-gt=0 at peak altitude => tpeak =Vyo/g so the time of flight is
2Vosin34.3/g
14.8=2*Vo^2*cos34.3*sin34.3 => Vo = 3.98m/s

. Since the shot velocity was 2.5 before being accelerated,

about 1.5ms were added by the force of the athlete.

The KE before the athlete pushed the shot is 0.5*6.85*2.5^2 and after is 0.5*6.85*4^2. The difference is 0.5*6.85[4^2-2.5^2=. 33.04J
That change in energy is the net force *1.65m which gives net force = 20N
20=F-6.85*9.8*sin34.3 => F=56.7N

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