How Do You Solve Incline and Spring Problems in Physics?

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The discussion revolves around solving incline and spring problems in physics, specifically involving an ice cube and a plastic cube on a slope. The ice cube, when released from a compressed spring, travels a distance of 51 cm up the slope before reversing direction, while the plastic cube, affected by friction, requires additional calculations to determine how far it travels. Participants emphasize the importance of incorporating friction into their equations, suggesting the use of force diagrams to find the normal force. The conversation highlights the need for a clear understanding of energy conservation and the role of kinetic friction in incline problems. Overall, the thread serves as a collaborative effort to clarify the steps necessary for tackling these types of physics questions.
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Homework Statement


A 50 g ice cube can slide without friction up and down a 30 degree slope. The ice cube is pressed against a spring at the bottom of the slope, compressing the spring 10 cm. The spring constant is 25 N/m. When the ice cube is released, what total distance will it travel up the slope before reversing direction?

b)The ice cube is replaced by a 50 g plastic cube whose coefficient of kinetic friction is 0.20. How far will the plastic cube travel up the slope?


Homework Equations


1/2mv^2
1/2kx^2


The Attempt at a Solution


I'm not a fan of incline problems and I honestly have no clue on how to start this question. Can someone please guide me through it and help me come up with a set of steps that will help me through future incline problems?

Homework Statement



The spring shown in the figure is compressed 50 cm and used to launch a 100 kg physics student. The track is frictionless until it starts up the incline. The student's coefficient of kinetic friction on the 30 degree incline is 0.15.

a)What is the student's speed just after losing contact with the spring?
b)How far up the incline does the student go?

http://img530.imageshack.us/img530/6145/knightfigure1154nl2.th.jpg

Homework Equations


1/2kx^2
1/2mv^2


The Attempt at a Solution


I got part a) which was 14.1m/s using the law of conservation of energy, but as I mentioned before, I always get stuck on incline questions.
 
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um, ok well i got the answer to part a of question 1 which was 51cm, but I am not sure how to include friction in the incline problem. can anyone please help?
 
\frac {1}{2}m[v^{2} - v_{0}^{2}] + mg[y - y_{0}] + \frac {1}{2}k[x^{2} - x_{0}^{2}] = W_{F} + W_{ext}

so now there is no spring involved and at the top, we stop moving further and eventually slide down. so we take that instant to be v=0, our equation simplifies to

\frac {1}{2}m[v^{2} - v_{0}^{2}] + mg[y - y_{0}] + = W_{F}

for part b
 
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i think i got confused

A 50 g ice cube can slide without friction up and down a 30 degree slope. The ice cube is pressed against a spring at the bottom of the slope, compressing the spring 10 cm. The spring constant is 25 N/m. When the ice cube is released, what total distance will it travel up the slope before reversing direction?

b)The ice cube is replaced by a 50 g plastic cube whose coefficient of kinetic friction is 0.20. How far will the plastic cube travel up the slope?

is there a picture by chance? what is y_initial?
 
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Draco said:
um, ok well i got the answer to part a of question 1 which was 51cm, but I am not sure how to include friction in the incline problem. can anyone please help?
in order to include friction, you must draw another force body diagram and solve for the Normal force

you're already given kinetic friction, so just find normal and you're good to go!

W_{F}=\mu_{k}N\cos{\phi}

\phi=\mbox{angle between}
 
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but with that do i add it to the potential energy, or only use the frictional force energy?
 
sorry, but there is only a picture for the second one. Part b of both questions are similar so the method would be the same right?
 
Draco said:
sorry, but there is only a picture for the second one. Part b of both questions are similar so the method would be the same right?
basically

and the equation is still the same, just modified ... and I'm not sure what y-final or y-initial is, but of course y-initial should be taken as 0 and y-final, idk; and I'm sure you're solving for s.

\frac {1}{2}m[v^{2} - v_{0}^{2}] + mg[y - y_{0}] + = \mu_{k}Ns\cos{\phi}

but in order to get N, you need a FBD ... which will most likely equal the weight.
 
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HOw did you get 51 cm for the first part?
 
  • #10
I didnt use a velocity, since there is only gravitational energy present in the equation. That's how i got 51. I know its right, my answer matches the correct answer
 
  • #11
I don't doubt that you are right. I just cant' seem to come to that value with my calculations.
 
  • #12
Er.. I am not sure i keep getting 78 as my answer for how much he goes up the incline.. my h was 0 since it was he was at 0 before he goes up the incline, and my v was 14.1m/s and i solved for s but i got it wrong
 
  • #13
I got part B to that question..
 
  • #14
a = -g(ukcos0 +sino)
 
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