WORK by Constant Force Argh help

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Homework Help Overview

The discussion revolves around a physics problem involving work done by a constant force, specifically a compressed spring launching a package up a frictionless ramp. The problem includes calculations related to the speed of the package upon reaching the truck and the effects of kinetic friction on subsequent packages.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of forces and energies involved, questioning the application of conservation of energy principles. There is an exploration of the relationship between spring force and kinetic energy, as well as the impact of friction on the package's ability to reach the truck.

Discussion Status

Some participants have provided guidance on using conservation of energy to analyze the problem, suggesting a focus on potential and kinetic energy. There is an ongoing exploration of the setup and assumptions, particularly regarding the role of friction and the configuration of the spring and ramp.

Contextual Notes

Participants note the absence of a figure that illustrates the setup, which may affect the clarity of the discussion. There is also mention of homework constraints that may limit the information provided.

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.
 

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