WORK by Constant Force Argh help

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The discussion revolves around calculating the speed of a package shot up a frictionless ramp using a compressed spring and addressing the impact of friction on the package's ability to reach the truck. For Part A, the user initially misapplies the spring force and kinetic energy equations, leading to incorrect calculations. The correct approach involves using conservation of energy principles, factoring in spring potential energy, gravitational potential energy, and kinetic energy. For Part B, the introduction of a sticky spot with friction raises concerns about whether the package can still reach the truck, which requires further analysis based on the energy lost to friction. The thread emphasizes the importance of correctly applying physics principles to solve the problem.
Kalie
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WORK by Constant Force...Argh...help

A freight company uses a compressed spring to shoot 1.80 kg packages up a 1.0-m-high frictionless ramp into a truck, as the figure shows. The spring constant is 337 N/m and the spring is compressed 34.0 cm.

A) What is the speed of the package when it reaches the truck?

B) A careless worker spills his soda on the ramp. This creates a 50-cm-long sticky spot with a coefficient of kinetic friction 0.30. Will the next package make it into the truck? Yes or No?

Okay...For Part A this is what I did but it was wrong

First I said F_s=-k x= 337(.34) = 114.38
with this I said that 114.38=1/2mv^2 v=11.28
The F_w=1/2mv^2=ma=1.8*-9.8=-17.64
W=F_w*D=F_w=17.64
W= the change in KE
KE_f=W+KE_i
KE_f=96.74
1/2mv^2=96.74
so v=10.36
That is wrong and I don't know how to get the right answer could someone help?

Part B will probably make more sense after I solve the first part but could someone help me with the setup?
 
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Kalie said:
Okay...For Part A this is what I did but it was wrong

First I said F_s=-k x= 337(.34) = 114.38
Here you calculate the force of the spring when it is at max compression.
with this I said that 114.38=1/2mv^2 v=11.28
Then you set this force equal to the KE! (Force and energy are two different things!)
The F_w=1/2mv^2=ma=1.8*-9.8=-17.64
W=F_w*D=F_w=17.64
W= the change in KE
KE_f=W+KE_i
KE_f=96.74
1/2mv^2=96.74
so v=10.36
That is wrong and I don't know how to get the right answer could
Try this: Consider conservation of energy. What's the energy when the spring is fully compressed and the package is at it's lowest point? (You didn't supply the figure, so it's not clear how the spring and ramp are connected. Is the spring on the ramp? Post the figure if you can.)
 
OK. Now use conservation of energy to find out how much KE the package has when it gets to the top of the ramp. Hint: Consider spring PE, gravitational PE, and KE.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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