Practice Test Questions

  • Thread starter Axbrown
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Homework Statement


A 10.0-kg block is released from rest on a quarter pipe 3.00m high. The track is frictionless except for a length (6.00m). The block travels down this track and hits a spring of force constant 2,250 N/m, and compresses the spring 0.300m from its equilibrium position before coming to rest momentarily. What is the coefficient of kinetic friction between the block and the rough surface on the 6.00m length?


Homework Equations


F=u*mg
Hooke's Law F=-k*deltax

The Attempt at a Solution



I plugged 2,250 N/m into Hooke's law equation along with .300m to get 675N, which I assumed was the force of the block going down the ramp. Then I plug mass, gravity, and force into the equation above F=u*mg with mass being 10kg and g being 9.81m/s/s and get 6.8 and I know the answer is .328. What am I doing wrong?
 

Answers and Replies

  • #2
BiGyElLoWhAt
Gold Member
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First of all there is no force of the block going down the ramp. As the block goes down the ramp, different forces are acting on it, but you don't really need to look at that. Try Conservation of Energy.
 
  • #3
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Okay so I am attempting to do this using Conservation of Energy, and I know my GPE is 294.3J, and using the information provided for the spring I found the spring's Mechanical Potential Energy to be 101.25J. Since that is the energy of the spring on the bottom and the block was able to compress it, does that mean the kinetic energy is also 101.25J?

Assuming it was, I subtracted my Kinetic energy from my GPE to get the energy lost which was 193.05J and divided it by the distance traveled to get 32.2J, which I believe is my energy lost due to friction. No I should be able to set up an equation and solve for μ, but I still don't have theta... as it is not, I have

f=μmgcos(theta)
32.2J=μ98.1kgm/s^2cos(theta)

If I divide both sides by 98.1kgm/s^2cos(theta) I should get μ, but once again I do not know theta. I'm having a hard time finding it since the slope is curved and not flat. I'm about to try and find it using Projectile motion, but let me know if I'm on the right track or if not give me some tips to steer me the correct way. Thanks

Ok so I just used KE=1/2*mv^2 to solve for velocity using KE found above (assuming it was correct) and then used velocity and the length of the ramp to find the degree angle above the ramp, and subtracted it from 90 to get the actual angle of the ramp which was 41.8 degrees. I plugged this into 32.2J=μ98.1kgm/s^2cos(41.8) and got μ=0.44. It's a lot closer than I've come to the answer (which is aμk=0.328) but its still not correct. Guidance would be greatly appreciated. thanks again
 
Last edited:
  • #4
billy_joule
Science Advisor
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The question is poorly worded. This is how I read it:
The quarter pipe is friction-less and 3 metres tall, the block slides down the quarter pipe then onto a horizontal track that is 6 metres long.
(Had I not been familiar with quarter pipes from skateboarding I may have interpreted it another way..)

So there is no cos theta term required in your equation:

32.2J=μ98.1kgm/s^2

that gives:
μ = 0.328
 
  • #5
BiGyElLoWhAt
Gold Member
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@Axbrown , you're good up until the friction portion and actually solving for mu,

I think billy_joule is right with his interpretation of the question, first of it's not a ramp it's a quarter pipe, second of all it's 3 m high and the section of ground with friction is 6m long. Try that and see what it gives.
 

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