How Is the Average Force Calculated During a Shot Put Acceleration?

[passphysicsC]

A shot putter launches a 7.260 kg shot by pushing it along a straight line of length 1.650 m and at an angle of 34.1 degrees from the the horizontal, accelerating the shot to the launch speed from its initial speed of 2.5 m/s (which is due to the athletes preliminary motion). The shot leaves the hand at a height of 2.110 m and at an angle of 34.1 degrees, and it lands at a horizontal distance of 15.90 m.

What is the magnitude of the average force on the shot during the acceleration phase?

 

[jbsteele]

The average force on the shot during the acceleration phase is equal to the change in momentum divided by the time taken for the acceleration phase. The change in momentum is equal to the final momentum of the shot minus the initial momentum. The final momentum of the shot can be calculated using its final velocity and mass, while the initial momentum is equal to zero since the initial velocity was 2.5 m/s. The time taken for the acceleration phase is equal to the time it takes the shot to travel 1.650 m. This can be found by dividing the distance traveled by the final velocity. Therefore, the magnitude of the average force on the shot during the acceleration phase is equal to: F = (m*vf - 0) / (d/vf) = (7.260 kg * 24.07 m/s - 0) / (1.650 m / 24.07 m/s) = 883.55 N