Join TSR now and get all your revision questions answeredSign up now

2 Potential Energy questions Watch

    • Thread Starter
    Offline

    0
    ReputationRep:
    Some of you may have noticed that I've been posting quite frequently for help with questions involving Potential/Kinetic Energy. It is a topic I've been struggling with a little, but this should be the last time for a while that I'll be needing help with it.

    1)
    A smooth wire is bent into the shape of the graph of  y=x+2 \sin x for  0<x<\pi , the units being metres. Points A, B and C on the wire have coordinates  (0,0), (\pi,\pi), (2\pi,2\pi) . A bead of mass m kg is projected along the wire from A with speed u ms^{-1} so that it has enough energy to reach B but not C. Prove that u is between 8.66 and 11.10.

    If u=10, the bead comes to rest at a point D between B and C. Find the greatest speed of the bead between B and D,

    AnswerSo I assume that if the bead has enough energy to reach B but not C, then it's fair to model that as mg2\pi>\frac {1}{2}mv^2>mg\pi. Doing the algebra gives me 2\sqrt{g\pi}>v>\sqrt{2g\pi} , giving me 11.10>v>7.85. So my upper value is correct, but not the lower one, and I'm quite confused as to why.

    For the next part, I assumed that the greatest speed of the bead between B and D would be the speed of the bead as it just passes B (as the graph of y=x+2\sin x always as a positive or flat gradient). The answer is apparantly  7.2 ms^{-1}, and I don't have any idea how they got that.


    So now for question 2
    2 A block of mass M is placed on a rough horizontal table. A string attached to the block runs horizontally to the edge of the table, passes round a smooth peg, and supports a sphere of mass m attached to its other end. The motion of the block on the table is resisted by a frictional force of magnitude F, where F<mg. The system is initally at rest.
    a) Show that when the block and the sphere have each moved a distance h, their common speed is given by v^2=\frac{2(mg-F)h}{M+m}
    b) Show that the total energy lost by the sphere as it falls through the distance h is \frac{m(Mg+F)h}{M+m}

    AnswerNow I can manage part a: Fs-Rs = Energy gain , so mgh-Fh = \frac {1}{2} (M+m)v^2. This is easy enough to rearrange to make v^2 the subject and get the expression in part a).

    Part b is giving me problems. I assume the energy lost is equal to  Fh, which is equal to mgh - \frac{1}{2}mv^2, but when I substitute in the answer for v^2 from part a) I just get a mess that no matter how I try to tidy up never looks anything like the expresssion I should be ending up with.

    Help would be extremely appreciated.
    Offline

    3
    ReputationRep:
    hmmm, I get 7.85 as well
    • Study Helper
    Offline

    3
    If you graph the curve, you'll see that in order to reach B, it must go over a hump earlier on, which is higher than B.

    Edit: If you don't want to sketch the curve, the fact that the gradient of the curve at B is negative, will also tell you that there is a hump prior to B.
    Offline

    3
    ReputationRep:
    (Original post by ghostwalker)
    If you graph the curve, you'll see that in order to reach B, it must go over a hump earlier on, which is higher than B.

    Edit: If you don't want to sketch the curve, the fact that the gradient of the curve at B is negative, will also tell you that there is a hump prior to B.
    Ahhhhhhh I differentiated and sketched x + sinx so missed that
    • Thread Starter
    Offline

    0
    ReputationRep:
    Can anyone help with the second one?
    Offline

    3
    ReputationRep:
    (Original post by Julii92)
    Can anyone help with the second one?
    Your start to 2b is correct and leads to the correct answer

    Energy = mgh - \dfrac{m(mg-F)h}{M+m}

    add fractions

    Energy = \dfrac{(mg(M+m)-m^2g+mF)h}{M+m}


    Energy = \dfrac{(mMg+m^2g-m^2g+mF)h}{M+m}


    ... ...
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by TenOfThem)
    Your start to 2b is correct and leads to the correct answer

    Energy = mgh - \dfrac{m(mg-F)h}{M+m}

    add fractions

    Energy = \dfrac{(mg(M+m)-m^2g+mF)h}{M+m}


    Energy = \dfrac{(mMg+m^2g-m^2g+mF)h}{M+m}


    ... ...
    Right! Thank you (And Ghostwalker)
 
 
 
Poll
Is GoT overrated?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Quick reply
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.