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2 Potential Energy questions

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+2sin⁡x y=x+2 \sin x for 0<x<π 0<x<\pi , the units being metres. Points A, B and C on the wire have coordinates (0,0),(π,π),(2π,2π) (0,0), (\pi,\pi), (2\pi,2\pi) . A bead of mass mm kg is projected along the wire from A with speed ums−1u ms^{-1} so that it has enough energy to reach B but not C. Prove that uu is between 8.668.66 and 11.1011.10.

If u=10u=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 mg2π>12mv2>mgπmg2\pi>\frac {1}{2}mv^2>mg\pi. Doing the algebra gives me 2gπ>v>2gπ2\sqrt{g\pi}>v>\sqrt{2g\pi} , giving me 11.10>v>7.8511.10>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+2sin⁡xy=x+2\sin x always as a positive or flat gradient). The answer is apparantly 7.2ms−1 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 MM 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 mm attached to its other end. The motion of the block on the table is resisted by a frictional force of magnitude FF, where F<mgF<mg. The system is initally at rest.
a) Show that when the block and the sphere have each moved a distance hh, their common speed is given by v2=2(mg−F)hM+mv^2=\frac{2(mg-F)h}{M+m}
b) Show that the total energy lost by the sphere as it falls through the distance hh is m(Mg+F)hM+m\frac{m(Mg+F)h}{M+m}

AnswerNow I can manage part a: Fs−Rs=EnergygainFs-Rs = Energy gain , so mgh−Fh=12(M+m)v2mgh-Fh = \frac {1}{2} (M+m)v^2. This is easy enough to rearrange to make v2v^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 Fh, which is equal to mgh−12mv2mgh - \frac{1}{2}mv^2, but when I substitute in the answer for v2v^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.
Reply 1
Avatar for TenOfThem
TenOfThem
18
hmmm, I get 7.85 as well
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.
(edited 11 years ago)
Reply 3
Avatar for TenOfThem
TenOfThem
18
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 :smile:
Reply 4
Avatar for Julii92
Julii92
OP
Can anyone help with the second one?
Reply 5
Avatar for TenOfThem
TenOfThem
18
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−m(mg−F)hM+mEnergy = mgh - \dfrac{m(mg-F)h}{M+m}

add fractions

Energy=(mg(M+m)−m2g+mF)hM+mEnergy = \dfrac{(mg(M+m)-m^2g+mF)h}{M+m}


Energy=(mMg+m2g−m2g+mF)hM+mEnergy = \dfrac{(mMg+m^2g-m^2g+mF)h}{M+m}


... ...
(edited 11 years ago)
Reply 6
Avatar for Julii92
Julii92
OP
Original post by TenOfThem
Your start to 2b is correct and leads to the correct answer

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

add fractions

Energy=(mg(M+m)−m2g+mF)hM+mEnergy = \dfrac{(mg(M+m)-m^2g+mF)h}{M+m}


Energy=(mMg+m2g−m2g+mF)hM+mEnergy = \dfrac{(mMg+m^2g-m^2g+mF)h}{M+m}


... ...


Right! Thank you :smile: (And Ghostwalker)

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