Dropping meat on weighing scale

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To determine the initial compression of a spring when a 1kg piece of meat is dropped onto a 2kg plate on a weighing scale, conservation of energy principles are applied. The potential energy lost by the meat as it falls is converted into elastic potential energy in the spring upon impact. The correct equation to use is 1/2mv^2 = 1/2kx^2, where m represents the total mass of the meat and plate, and v is the velocity just after impact. It's important to avoid double counting the potential energy, as the kinetic energy already accounts for the loss in potential energy. This approach allows for accurate calculation of the spring's initial compression due to the impact.
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hi, if we drop a piece of meat say 1kg from 1m above the plate, on a plate 2kg, and the plate is on a weighing scale of spring F=kx, how do we find how much the spring will first compress? ( not referring to the equilibrium where total weight = spring force)

so we can't use F=kx = mg here because the mg is for the piece of meat only, but how do i take into account the plate?


and even without the plate, when i first drop the meat from height 1m above the scale, the spring will still compress more than when it reaches equilibrium (mg=kx). so how do i find this initial compression ?

thanks
 
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Using conservation of energy,

The loss in Potential energy of the meat and plate as the spring compress = gain in elastic potential energy of the spring.
 
quietrain said:
so we can't use F=kx = mg here because the mg is for the piece of meat only, but how do i take into account the plate?
In the case of two objects (meat and plate) that impact and don't bounce apart, you can use conservation of energy and also the conservation of momentum to determine the velocity of the pair immediately after impact. Once they impact and move together with a new initial velocity, you can then apply conservation of energy to the pair of objects and compare it to the spring energy.
 
A good engineering rule of thumb is that a dynamic strain is twice the static strain of the same load. Your load will be even higher than that since you are not only applying the strain dynamically, but with some initial kinetic energy.
 
ok, so my equation is now mgh + 1/2mv^2 = 1/2 kx^2

but my m in mgh is the meat only? and h is the height above the plate?

my m in KE is meat + plate? v is final speed determined from conservation of momentum?

and so my x is the initial compression due to the impact?
 
the mgh in ur context is wrong. You are double counting the loss in PE since you already took into account the Kinetic energy arised from the loss in PE
 
oh. so my equation should be just 1/2mv^2 = 1/2kx^2 ?

where m is the total mass ??
 
Looks right to me.
And v is the speed just after the meat lands (1/3 of the meat speed).
 
ah ok thanks everyone
 
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