Max height at the top of the ramp of object launched by spring

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SUMMARY

The discussion centers on a physics problem involving a 2 kg block launched by a spring with a spring constant of 1000 N/m on a frictionless horizontal platform leading to a ramp with a kinetic friction coefficient of 0.3 and an incline of 20 degrees. The user calculated the initial kinetic energy from the spring to be 20 J and attempted to find the height at the top of the ramp using the conservation of energy principle. The correct approach involves calculating the work done against friction and using trigonometry to relate the height to the distance traveled along the ramp. The final height calculated was approximately 0.7385 meters, which requires verification of the frictional work done.

PREREQUISITES
  • Understanding of conservation of energy principles in physics
  • Knowledge of kinetic and potential energy calculations
  • Familiarity with frictional force calculations, including Ffric = μFN
  • Basic trigonometry to relate ramp height and distance
NEXT STEPS
  • Review the concept of elastic potential energy and its conversion to kinetic energy
  • Learn how to calculate work done by friction in inclined planes
  • Study the relationship between forces on inclined surfaces, including normal force calculations
  • Explore more complex problems involving energy conservation in multi-step scenarios
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Students studying physics, particularly those focusing on mechanics and energy conservation, as well as educators looking for examples of real-world applications of these concepts.

andrewwg94
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Homework Statement


Ok so I had this problem on my quiz today and I am really anxious to see if i got it right.
It went something like this:

You have a frictionless horizontal platform and a ramp with kinetic friction of 0.3. The ramp is 20 degree of the ground. A 2 kg block/object is launched by a spring with k=1000 N/m at the beginning of the horizontal ramp. The spring is stretched back 20 cm. So the idea is that it is launched on the horizontal platform before eventually going up the ramp. use 9.8 for g.


Homework Equations



I used KEi + PEi - Wfriction = KEf+PEf

The Attempt at a Solution


I basically found the KE due to the spring to be 20N. Then using that i calculated work lost due to friction. P. Energy at top of ramp = mgh. so i solved for h.

I got h=.7385

Everything made sense when i did it. I don't know if that's right though.
 
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Welcome to PF,

andrewwg94 said:
I basically found the KE due to the spring to be 20N. Then using that i calculated work lost due to friction. P. Energy at top of ramp = mgh. so i solved for h.

Did you mean 20 J (joules)? Energy is measured in joules, not Newtons. If so, this is correct. Your equation, which expresses the basic physical concept behind this problem (namely the conservation of energy) is also correct. Energy is conserved, hence:

KEi + PEi = KEf + PEf + Wfric

Initially, the object is not moving, hence KEi = 0. All of the energy is in the form of elastic potential energy (energy stored in the spring). So PEi = (1/2)kx2 = 20 J.

At the very end, when the object has traveled as high as it is going to go, it is also not moving. Hence, KEf = 0. All of these 20 joules have been converted into some combination of gravitational potential energy and heat (the latter being equal to the work done by friction).

20 J = mgh + Wfric

Wfric = Ffric * d

where Ffric is the frictional force, and d is the distance traveled along the ramp by the object. The distance d is related to h by simple trigonometry, which means that you can express everything in terms of one variable: h.

For the frictional force, Ffric = μFN, where FN is the normal force acting on the object. The normal force always acts perpendicular (i.e. normal) to the two surfaces that are in contact. So, in this case, the normal force is perpendicular to the ramp surface, which means that you need to use the geometry (specifically the inclination angle of the ramp) to figure out what it is. You can figure out what it is by noting that all forces perpendicular to the ramp have to balance each other out, so the normal force is just equal (in magnitude) to the component of the weight that acts perpendicular to the ramp.

So, it looks like you may have been doing well in the beginning, but then did not compute Wfric properly, at least based on what you posted above.
 

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