What is the average force exerted by a punter on a football in Newton's Law?

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SUMMARY

The average force exerted by a punter on a football can be calculated using Newton's second law of motion. In this discussion, a punter accelerates a 0.50 kg football from rest to a speed of 10 m/s over a contact time of 0.18 seconds, resulting in an average force of approximately 27.78 N. Additionally, the discussion covers a freight train scenario where a force of 6.8 x 10^5 N is used to accelerate a train with a mass of 1.3 x 10^7 kg from rest to 84 km/h, demonstrating the application of force and acceleration calculations.

PREREQUISITES
  • Understanding of Newton's second law of motion
  • Basic knowledge of force, mass, and acceleration
  • Ability to convert units (e.g., km/h to m/s)
  • Familiarity with kinematic equations
NEXT STEPS
  • Study the derivation and applications of Newton's second law of motion
  • Learn how to convert between different units of speed and acceleration
  • Explore kinematic equations for uniformly accelerated motion
  • Investigate real-world applications of force calculations in sports and transportation
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Students studying physics, particularly those focusing on mechanics, as well as educators seeking to clarify concepts related to force and motion.

BunDa4Th
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Well, I am having trouble with this and I been looking at the book for a long time and still lost. The teacher never went over this so no notes to help me out here either.

A football punter accelerates a football from rest to a speed of 10 m/s during the time in which his toe is in contact with the ball (about 0.18 s). If the football has a mass of 0.50 kg, what average force does the punter exert on the ball?
N

I have no slight clue on where to start or what formula I should be using. Any help would be great.
 
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I think that means the football's starting speed is 10m/s which is the time the football punter kick it!
 
edit: okay i figure out how to do the first one but at a lost on this problem

A freight train has a mass of 1.3 x 10^7 kg. If the locomotive can exert a constant pull of
6.8 x 10^5 N, how long does it take to increase the speed of the train from rest to 84 km/h? min

so far i got this part from reading the book ma = F_net --> a = F_net/m =

6.8 x 10^5 N/1.3 x 10^7 kg = .052 m/s^2

84km/h x .278

I think i use v = at + V_0 = (.052)t = 23.352 t = 449 second

edit: okay nevermind i figure this out do to my own mistake. I got the wrong conversion for km/h to m/s. I done correction to it.
 
Last edited:

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