- #1
Jazz
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Homework Statement
Very large forces are produced in joints when a person jumps from some height to the ground. Calculate the magnitude of the force produced if an ##\small{80.0\!-\!kg}## person jumps from a ##\small{0.600\!–\!m\!-\!}##high ledge and lands stiffly, compressing joint material ##\small{1.50\ cm}## as a result. (Be certain to include the weight of the person).
Known data:
##m = 80.0\ kg##
##h = 0.600\ m##
##d = 1.50 \times 10^{-2}\ m##
Homework Equations
##W_{nc} = \Delta KE + \Delta PE##
The Attempt at a Solution
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The work done is negative since it removes mechanical energy from the system:
##- W_{nc} = KE_f - KE_i + PE_f - PE_i##
##- W_{nc} = - PE_i##
##F_{nc}d = mgh##
##F_{nc} = \frac{mgh}{d} = \frac{(80.0\ kg)(9.8\ m/s^2)(0.600\ m)}{1.50 \times 10^{-2}\ m}##
##F_{nc} = 3.13 \times 10^4\ N##
Here is where I thought I was done. The problem asks to include the weight of the person. By doing so, I get ##\scriptsize{3.21 \times 10^4\ N}##, which agrees with the answer given to it, but I've not been able to see what is the point of adding their weight.
As fas as I understand the weight is already being included in ##F_{nc}##.
If I think about it in terms of compression, when standing their ##\scriptsize{(80.0\ kg)(9.8\ m/s^2) = 784\ N}## are always compressing the joints a little bit.
If I added a barbell weighing ##\frac{3.134 \times 10^4\ N - 784\ N}{9.8\ m/s^2} = \scriptsize{3.12 \times 10^3\ kg}## on top of the man, it would be enough to compress their joints 1.50 cm.
Asking me to add the weight of the man (again, from my view) it's like saying that ##F_{nc}## does not include the man's weight (I can't see why not) and, when standing, the man's weight doesn't compress the joints at all.
Thanks!
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