Solving Dave's Cardboard Contact Force Problem

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Dave's problem involves calculating how much cardboard boxes contract when he lands after falling at a velocity of 2 m/s. The force exerted by the boxes is 350 N, while Dave's weight is 70 kg. The discussion highlights confusion over using delta y versus delta x in kinematic equations, with the consensus that both can be interchangeable in this context. It emphasizes the importance of understanding the correct application of kinematic equations and the role of gravity in the calculations. Ultimately, the key to solving the problem lies in properly integrating the force and mass with the appropriate equations.
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Hi this is a problem from my midterm at the beginning of the quarter and i have a final coming up so i need to know the previous midterm material but i still don't understand this one question.

Dave suddenly drops from the wire and lands at a velocity of 2m/s on some cardboard boxes that contract to absorb the force of his fall. If the cardboard boxes exerts a force of 350N on Dave who weighs 70kg, how much do the cardboard boxes contract?


the equation i would use is v^2=vo^2 - 2adelta y

but in her solutions she shoes delta x. Why is it delta x if the cardboard is going to be contracting in the y direction?

If someone can help me start off this problem i will so grateful!
 
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Off the top of my mind, I think her equations are supposed to be general. You can substitute delta y for delta x. If you continue on with that assumption, does it get you the right answer?
 
yeah that's what i thought too but there are two equations, one for x and one for y and they differ.

v^2 = Vo^2+2adelta x

v^2 = Vo^2-2gdelta y

in x you add, in y you subtract
 
That's general kinematics. You can't just plug in the exact numbers for the exact equations and hope for the best. At what rate is David accelerating when he jumps off? I assume that his initial vertical v is zero. If you figure that out, you can find the unknown y.
 
nvm i think youre right. It was just difficult for me to apply the force and the mass to the equation. I thought gravity would be used instead.
 
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