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    So we got

    Particle A is going down a slope of 14m. At the start, it was at 12 ms-1 and at the end it was 8 ms-1.

    The mass of A is 3.5kg, and the slope is angled at 20 degrees.

    Work out the friction coefficient.

    How would I do this?????
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    (Original post by RickHendricks)
    So we got

    Particle A is going down a slope of 14m. At the start, it was at 12 ms-1 and at the end it was 8 ms-1.

    The mass of A is 3.5kg, and the slope is angled at 30 degrees.

    Work out the friction coefficient.

    How would I do this?????
    Draw a diagram and label forces. There will be a frictional force 3.5g\mu \cos 30 and a weight force 3.5g \sin 30, both parallel to the slope, which make up your F=ma. You can hence determine a in terms of \mu. The acceleration is constant, so you can use v^2=u^2+2as to determine \mu.
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    (Original post by RickHendricks)
    So we got

    Particle A is going down a slope of 14m. At the start, it was at 12 ms-1 and at the end it was 8 ms-1.

    The mass of A is 3.5kg, and the slope is angled at 30 degrees.

    Work out the friction coefficient.

    How would I do this?????
    I'd probably use energy. Have you learnt the work/energy principle?
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    (Original post by Notnek)
    I'd probably use energy. Have you learnt the work/energy principle?
    yes. That was the previous question.

    I worked out the work done as 140J
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    (Original post by RickHendricks)
    yes. That was the previous question.

    I worked out the work done as 140J
    Can you please post what you've tried for this question?
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    (Original post by Notnek)
    Can you please post what you've tried for this question?
    yes

    Energy lost = work done = change in Kinetic Energy

    Change in kinetic energy = (1/2 x 3.5 x 12^2) - (1/2 x 3.5 x 8^2)
    which gave me 140J

    I don't see now how I can use this. I can work out the force of the particle going down the slope, which would be 140/14 (since 14 is the displacement, and using the formula work done = force x displacement)

    Then this means that Fmax would be 10, since it's decelerating, meaning that friction is at it's maximum? Is it right doe? This is the part where I get stuck.
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    (Original post by RickHendricks)
    yes

    Energy lost = work done = change in Kinetic Energy

    Change in kinetic energy = (1/2 x 3.5 x 12^2) - (1/2 x 3.5 x 8^2)
    which gave me 140J

    I don't see now how I can use this. I can work out the force of the particle going down the slope, which would be 140/14 (since 14 is the displacement, and using the formula work done = force x displacement)

    Then this means that Fmax would be 10, since it's decelerating, meaning that friction is at it's maximum? Is it right doe? This is the part where I get stuck.
    You haven't considered the gain in gravitational potential energy:

    Energy lost = Loss in KE - Gain in PE

    EDIT: Oops made a mistake here.
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    (Original post by RickHendricks)
    Then this means that Fmax would be 10, since it's decelerating, meaning that friction is at it's maximum? Is it right doe? This is the part where I get stuck.
    Since the particle is moving (acceleration is irrelevant), the friction will be at its maximum.

    F_{max} = \mu R

    So you need to find the reaction of the slope on the particle.

    Then use work done against friction = F_{max}\times 14.
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    (Original post by Notnek)
    You haven't considered the gain in gravitational potential energy:

    Energy lost = Loss in KE - Gain in PE
    I would need the height for that. I haven't been given that,
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    (Original post by RickHendricks)
    yes

    Energy lost = work done = change in Kinetic Energy

    Change in kinetic energy = (1/2 x 3.5 x 12^2) - (1/2 x 3.5 x 8^2)
    which gave me 140J

    I don't see now how I can use this. I can work out the force of the particle going down the slope, which would be 140/14 (since 14 is the displacement, and using the formula work done = force x displacement)

    Then this means that Fmax would be 10, since it's decelerating, meaning that friction is at it's maximum? Is it right doe? This is the part where I get stuck.
    KE before + mgh = KE after + work done against friction
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    (Original post by RickHendricks)
    I would need the height for that. I haven't been given that,
    You know it's travelled 14m so you can use trig with the angle to find the change in height.

    Have you practiced basic questions before trying this one? It sounds like you're missing some of the basics.
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    (Original post by Notnek)
    You haven't considered the gain in gravitational potential energy:

    Energy lost = Loss in KE - Gain in PE
    energy lost has to be positive right?

    My loss in KE was 140J, and my gain in PE, by using P.E = mgh, where h is sin30x14, i get gain in P.E as (49/2)g, which means that as 140 - (49/2)g would be energy lost, which comes out negative.

    I dont think I get this topic very well. Sorry. However isn't the energy loss meant to be positive?
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    (Original post by RickHendricks)
    energy lost has to be positive right?

    My loss in KE was 140J, and my gain in PE, by using P.E = mgh, where h is sin30x14, i get gain in P.E as (49/2)g, which means that as 140 - (49/2)g would be energy lost, which comes out negative.

    I dont think I get this topic very well. Sorry. However isn't the energy loss meant to be positive?
    How did you get 140 as loss in KE? Can you please show working for this?

    It is a confusing topic at first but I'm sure with practice you'll be fine.
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    (Original post by Notnek)
    You know it's travelled 14m so you can use trig with the angle to find the change in height.

    Have you practiced basic questions before trying this one? It sounds like you're missing some of the basics.
    hang on, found a mistake somewhere. I'll reply back when I fix it all.
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    (Original post by Notnek)
    How did you get 140 as loss in KE? Can you please show working for this?

    It is a confusing topic at first but I'm sure with practice you'll be fine.
    ok for some reason the angle I put down was 30 degrees, and it was actually 20 degrees.

    So from the actual angle,

    total energy loss = K.E lost - P.E gained

    which goes to ((1/2 x 3.5 x 12^2)-(1/2 x 3.5 x 8^2))-3.5 x 9.8 x 14sin20

    this however gives me the answer of -24.2 (3 s.f), which is defintely wrong.

    The original answer is 304, so i don't know what I'm doing wrong.
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    (Original post by RickHendricks)
    ok for some reason the angle I put down was 30 degrees, and it was actually 20 degrees.

    So from the actual angle,

    total energy loss = K.E lost - P.E gained

    which goes to ((1/2 x 3.5 x 12^2)-(1/2 x 3.5 x 8^2))-3.5 x 9.8 x 14sin20

    this however gives me the answer of -24.2 (3 s.f), which is defintely wrong.

    The original answer is 304, so i don't know what I'm doing wrong.
    Sorry I made a mistake in my post earlier. The particle is travelling down the slope so it is losing GPE. So

    Total energy lost = loss in KE + loss in PE
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    (Original post by Notnek)
    Sorry I made a mistake in my post earlier. The particle is travelling down the slope so it is losing GPE. So

    Total energy lost = loss in KE + loss in PE
    ok that cleared all my working outs and gave me the correct answer

    since it's loss in KE + loss in PE, Total energy loss is 304J (3 s.f), and carrying that value over to the formula F = W.D x S, where F is Fmax, since it's moving, we get Fmax as 21.73....

    This means that using trig, we can get R as 3.5gCos20, since R is the same as the weigh (as it's same position horizontally), so by using those 2 values, I get Mu as 0.674, which is correct.

    Thanks!
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    (Original post by RickHendricks)
    ok that cleared all my working outs and gave me the correct answer

    since it's loss in KE + loss in PE, Total energy loss is 304J (3 s.f), and carrying that value over to the formula F = W.D x S, where F is Fmax, since it's moving, we get Fmax as 21.73....

    This means that using trig, we can get R as 3.5gCos20, since R is the same as the weigh (as it's same position horizontally), so by using those 2 values, I get Mu as 0.674, which is correct.

    Thanks!


    If you use this approach for work/energy questions then always analyse it first to check which types of energy are being lossed/gained then use

    Total loss in energy = sum of all the losses - sum of all the gains

    Total gain in energy = sum of all the gains - sum of all the losses


    There are other approaches that are worth knowing about if you find you're struggling with this. E.g. a formula that considers before/after:

    PE before + KE before + Work in - Work out = PE after + KE after
 
 
 
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