How Far Can a Greased Watermelon Be Pulled Down an Inclined Plane?

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SUMMARY

The discussion focuses on calculating the maximum distance a greased watermelon can be pulled down an inclined plane before the spring pulls it back. The watermelon has a mass of 4.72 kg, is on a ramp inclined at 27.4°, and has coefficients of static and kinetic friction of 0.560 and 0.392, respectively. The spring constant is 175.2 N/m. The final calculated distance is 0.253 meters after correcting the direction of the friction force, which opposes the motion of the watermelon.

PREREQUISITES
  • Understanding of Newton's second law (F=ma)
  • Knowledge of static and kinetic friction coefficients
  • Familiarity with spring mechanics and Hooke's Law
  • Basic trigonometry for resolving forces on an inclined plane
NEXT STEPS
  • Study the effects of different coefficients of friction on inclined planes
  • Learn about the dynamics of spring systems in physics
  • Explore advanced applications of Newton's laws in real-world scenarios
  • Investigate the role of angles in force resolution on inclined surfaces
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Students in physics, particularly those studying mechanics, as well as educators looking for practical examples of inclined planes and friction in action.

Jpyhsics
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Homework Statement


As part of an elaborate prank, a well-greased watermelon (m=4.72 kg) is placed on a ramp inclined with angle θ=27.4° with respect to the horizontal. Because of the grease, the coefficients of static and kinetic friction are µs= 0.560 and µk= 0.392 respectively. The watermelon is attached to the wall, parallel to the ramp, by a spring with constant k=175.2 N/m. What is the farthest distance down the ramp that the watermelon can be pulled away from the spring equilibrium and placed at rest on the surface without the spring pulling the watermelon back up the ramp? Express your answer in metres

Homework Equations


F=ma

The Attempt at a Solution


ΣFx=mgSinθ-kx-μsmgCosθ

Rearranged to make:
x=(mg(μsCosθ-Sinθ))/(-k)
x=-0.00977m

I used the coefficient of static friction, because it was at rest. Does my solution look correct?
I have included an image of my solution as well.
 

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Hello.

Does a negative answer for x seem right?

Check to see if you have the correct direction for the friction force.
 
TSny said:
Hello.

Does a negative answer for x seem right?

Check to see if you have the correct direction for the friction force.
Why would force of friction be towards the right?
 
Jpyhsics said:
Why would force of friction be towards the right?
According to the conditions stated in the problem, is the watermelon on the verge of slipping up the slope or slipping down the slope?
 
TSny said:
According to the conditions stated in the problem, is the watermelon on the verge of slipping up the slope or slipping down the slope?
Slipping up...oh so the friction has to oppose that verge of motion? My revised answer is 0.253m Does that seem right? And thank you so much!
 
Jpyhsics said:
Slipping up...oh so the friction has to oppose that verge of motion? My revised answer is 0.253m Does that seem right? And thank you so much!
Looks right to me.
 

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