3A
DL Quiz SEC _ _ Grade:
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04/26/10
First three letters of your last name
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Useful formulae: ETOT= Q + W; Eth= mcT; Ebond= |mH|; PEgrav= mgh; KEtr= mv2; PEms= kx2 Useful constants: g=9.8 m/s2
1. A skier skis down a mountain slope as shown below, beginning with a certain speed vi. Suppose you can entirely neglect the effects of friction and air resistance. a) Can you change the speed at the bottom by starting from a higher spot (without changing m or )? If so, how? If not, why not? Explain. You will be graded on the completeness and the quality of your explanation. m h Yes, you can. PE + KE = 0 mgh + 1/2m(vf2 vi2) = 0 , but vi remains unchanged. So, the bigger h, the bigger vf.
b) If the 70 kg skier is then initially at rest, and begins skiing down the slope an from initial height of 200 m, determine the speed of the skier at the bottom of the slope. gh + 1/2(vf2 vi2) = 0 vf = sqrt[-2gDh] = sqrt[(2)(10)(200)] = sqrt[4000] 63 m/s 2. The potential energy graph shown here is similar in shape to the 2-d Higgs potential in particle physics. The horizontal axis here is given in meters and the vertical axis in Joules. a) At which point (B or C) would the force be greatest? Clearly explain your reasoning. Point C would have a greater force than point B. Force is related to slope, and the slope at B=0 whereas the slope at C is >0. [Note that it doesnt matter which direction you think the force is (pos or neg slopes), just that the magnitude is bigger.]
A B
C
b) In which direction is the force at point A? Explain.
The force at point A is to the right. Force is always in the direction to locally decrease the PE.