One Dimensional Kinematics: Force

[TG3]

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

 

[Delphi51]

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.

 

[TG3]

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?

 

[Delphi51]

Oh dear, the computer is right! I forgot about the weight. Sorry.