Physics 218 Exam 2 Spring 2011, Sections , 526, 528

Transcription

1 Physics 8 Exa Sprin, Sections 53-55, 56, 58 Do not fill out the inforation below until instructed to do so! Nae Sinature Student ID E- ail Section # : Solutions in RED Rules of the exa:. ou have the full class period to coplete the exa.. When calculatin nuerical values, be sure to keep track of units. 3. ou ay use this exa or coe up front for scratch paper. 4. Be sure to put a box around your final answers and clearly indicate your work to your rader. 5. Clearly erase any unwanted arks. No credit will be iven if we can t fiure out which answer you are choosin, or which answer you want us to consider. 6. Partial credit can be iven only if your work is clearly explained and labeled. 7. All work ust be shown to et credit for the answer arked. If the answer arked does not obviously follow fro the shown work, even if the answer is correct, you will not et credit for the answer. Put your initials here after readin the above instructions:

2 Part : (5p) Forces and Dynaics Proble.: A trianular shape object of ass and aperture anle is between a wall and square block of ass as shown in the fiure below. There is friction between the wall and the trianular shape object with coefficient uk, and there is no friction anywhere else. As the trianular shape object wede its way down the µk block ets displaced in the horizontal position. Question..: (5p) As a first step, draw a coordinate syste and based on the eoetry derive an expression between the vertical acceleration of the trianular object and the horizontal acceleration of the square block. It is clear that!x-!y, " #a y Table to be filled by the raders Part Score Part (5) Part (5) Part 3 (5) Part 3 (5) Bonus () Exa Total Question..: (5p) Draw the free body diara of both the trianular shape object and the square block, properly indicatin relevant anles. Fk N3 N Nw N w Question..3: (5p) Usin Newton s laws find the acceleration of the square block in the horizontal direction. Fro the free body diara's above we obtain: ˆx : N w! N " N w N ŷ : F k! + N sin(!) a y " N! + N sin(!) a y! ˆx : N ŷ : N 3! N sin(!)! Where we've used!a y in the nd equation (riht). We are left with a syste of 4 equations and 4 unknown variables (N, N 3, N w, ). Replacin N w and N fro the first and third equations into the second equation we et: w a! +! x " µ a + a + k x x # + + % ( " $ ' ( + )+ ( + )+

3 Part : (5p) Equilibriu Proble.: A block of ass is hanin fro two ropes that are connected to a sinle rope, which in turn is attached to the wall as shown in the fiure below. The anles of the ropes, β, 9 are shown in the picture and are known. β Question..: (5p) very clearly draw the relevant free body diaras includin relevant anles and force naes. 3- rope point: block: T3 β T T Nw T T w Question..: (5p) Find the tension in the rope at an anle as a function of known paraeters. T +T sin(!)! N W!T " N W T!T 3 sin(") " T 3 sin(") T 3 cos(")!t!t sin(!) Pluin the last equation into the first one we et: T " T sin(") cos(")!t #!T sin(!) " T % $ tan(")! sin(!) ( T ' # % $ tan(")! sin(!) # (+T sin(!)! " T %! sin(!)+ sin(!) ( " T tan(") ' $ tan(") ' Question..3: (5p) Find the noral force that the wall needs to withstand so that the block does not sink into it. Fro the previous point we obtained N W,therefore N w tan(")! N w tan(") 3

4 Part 3: (5p) Work and Enery Proble 3.: A block of ass is at the botto of an inclined plane of lenth d and anle. The block is bein pulled up by a special winch located at the top of the plane and at a distance L fro the top of the plane as shown in the diara below. There is friction with coefficient of kinetic friction only in the inclined portion of the plane, and no friction at the top of the plane. The winch provides a constant force T to the rope and will work until the block reaches the winch. L Question 3..: (5p) Find the work of the tension force on the block done in brinin it all the way up to the winch. d Since the rope is always in the direction of the displaceent we et ro the previous point we obtained N W,therefore W T Td +TL T(d + L) Question 3..: (5p) Find the work of the friction force on the block done in brinin it all the way up to the winch. Since the friction force is always aainst direction of the displaceent and we only have friction force at the inclined portion of the plane we et W friction!fd! Nd. The noral force is obtained fro a free body diara a N so we et W friction!fd! d. Question 3..3: (5p) Usin the work enery theore find the velocity of the block at the oent it hits the winch. Usin a coordinate syste in which the axis oes up fro zero at the round we et: W T +W friction h + v f T(d + L)! µ d d sin(!)+ v f k " v f T(d + L)! d! d sin(!) " v f " v f T(d + L)! d (!sin(!) ) T(d + L)! d! d sin(!) 4

5 Part 4: (5p) Down the rap Proble 4.: A skier is at the top of a circular hill of radius R with horizontal velocity v. Question 4..: (5p) Knowin that there is no friction use the R v r work enery theore to express the velocity of the skier as a function of R, v and the anle of the skier. Usin a coordinate syste with vertical axis oin up and zero at the center of the circle R+ v v R+! v R+ v " R v + R(" )! v v + R(" ) Skier: r N Question 4..: (3p) Write a useful coordinate syste and the free- body diara of the skier with relevant anles. Question 4..3: (p) In ter of forces what happens when the skier looses contact with the hill? The noral force is zero. Question 4..4: (p) Find the anle at which the skier looses contact with the hill as a function of, R and v. Usin the coordinate syste depicted above we et in the ˆr direction: N!! v R At the anle! ' in which the skier lose contact with the round we have N, and therefore Rcos(! ') v and usin the v obtained fro conservation of enery we et Rcos(! ') v + R(! cos(! ')) " Rcos(! ')+ Rcos(! ') v + R " 3Rcos(! ') v + R " cos(! ') v 3R + 3 W Question 4..5: (5p) What is the inial velocity v at which the skier stops touchin the round even at the top of the hill? If the skier is to stop touchin the round alost at the top of the hill then! and therefore. Pluin this into the previous equation we et: cos(! ') v 3R + 3! v # 3R " % ( R! v R $ 3' 5