One Dimensional Kinematics: Force

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Homework Help Overview

The discussion revolves around a one-dimensional kinematics problem involving a ball's interaction with the floor after being dropped. The original poster presents a scenario where a ball is compressed upon impact, and they seek to determine the force exerted on the floor, assuming constant acceleration.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to use kinematic equations to find acceleration and subsequently the force. Another participant suggests an alternative approach using a velocity versus time graph to derive the same result. There is also a discussion about the inclusion of gravitational force in the calculations.

Discussion Status

Participants are exploring different methods to solve the problem, with some providing alternative reasoning. There is acknowledgment of the need to consider gravitational force in the final answer, indicating a productive direction in the discussion.

Contextual Notes

Participants question the assumption of constant acceleration and the implications of including gravitational force in the calculations. There is a mention of a computer system providing feedback on the correctness of the answers, which adds a layer of complexity to the discussion.

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Homework Statement



A .3 kg ball is compressed a maximum of 0.6 cm when it strikes the floor at 9.29 m/s. Assuming acceleration is constant, what is the force the ball exerts on the floor?

Homework Equations



vf^2 = v0^2 + 2A(x-x0)
Once I find A it will be easy, since
F=MA

The Attempt at a Solution



0^2 = 9.29^2 + 2A (.006)
0 = 86.3041 + .012 A
-86.3041= .012A
-7192= A
F=MA
F=.3 (-7192)
F= -2157
 
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That looks good to me! You have assumed constant acceleration, which probably isn't really right but which is probably a standard assumption in your course.

I actually did it a different way. I made a sketch of a v vs t graph, a straight line going from 9.29 at time 0 to zero at time t. The area under a v vs t graph is the distance .006. Using the area formula I was able to find the time t it takes for the ball to compress and stop. Then I used the idea that the slope on the v vs t graph is the acceleration. I got the same answer you have.
 
I found the "correct" answer: the computer wanted me to add the force due to gravity (.3 x 9.81) to the force exerted by the floor. This seems a bit conceptually shaky to me, but the computer said that was the correct answer. For my own future knowledge: is it, or was my first answer correct?
 
Oh dear, the computer is right! I forgot about the weight. Sorry.
 

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