Physics 218: Exam 2. Sections 501 to 506, 522, 524, and 526. March 8th, You have the full class period to complete the exam.

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1 Physics 18: Exam Sections 501 to 506, 5, 54, and 56. March 8th, 013. ules of the exam: 1. You have the full class period to complete the exam.. Formulae are provided on the last page. You may NOT use any other formula sheet. 3. When calculating numerical values, be sure to keep track of units. 4. You may use this exam or come up front for scratch paper. 5. Be sure to put a box around your final answers and clearly indicate your work to your grader. 6. Clearly erase any unwanted marks. No credit will be given if we cant figure out which answer you are choosing, or which answer you want us to consider. 7. Partial credit can be given only if your work is clearly explained and labeled. 8. All work must be shown to get credit for the answer marked. If the answer marked does not obviously follow from the shown work, even if the answer is correct, you will not get credit for the answer. Sign below to indicate your understanding of the above rules. Name : Student ID : Signature :

2 Part 1: Force and Energy Question Points Score Force and Energy 4 Cart 6 Funnel 4 Pulleys 30 Skiing 16 Total: 10 (4 points) An object of mass M = Kg is moving in the (x,y) plane under the influence of a force with potential given by U(x, y) = 1x J + 4y J + xy J. m m m (a) (8 points) Find both componentes of the force vector when the position of the object is (x=3m,y=4m). F = ( du dx, du ) = ( 1 y, 4 x) dy F (3, 4) = ( 1 4, 4 3)N F (3, 4) = ( 9, 10)N (b) (8 points) Find the single point of equilibrium of this potential. These are the points in which the force is zero. F = ( du dx, du ) = ( 1 y, 4 x) = (0, 0) dy from which we get (x 0, y 0 ) = (, 1 )m

3 (c) (8 points) It is noticed that when the object is at position (5,3) the speed is 8 m. Using conservation of energy, find the speed of the object if it reaches s position (1,1). K I + U i + W Other = K f + U f 1 Mv i + U(5, 3) = 1 Mv f + U(1, 1) vf = vi + (U(5, 3) U(1, 1)) M v f = 64( m s ) + (47 7)(m s ) v f = 104( m s ) (d) (8 points (bonus)) Find a relationship between the x-coordinate and y- coordinate of those points in the plane in which the force vector points in the same direction as the force at (1,1). [Hint: consider the angle as related to Fy F x ] At position (1,1) the vector points with angle α such tan(α) = F y (1, 1) = 4 + x F x 1 + y x=1,y=1 = 6 3 = The vectors that point with the same angle should have F y (x, y) = 4 + x F x 1 + y = 4 + x = + 4y y = 1 + x Page

4 Part : Cart (6 points) A cart, initially at rest, contains a spring and a block of mass m as depicted in the figure below. The coefficients of friction between the wall of the cart and the block are µ k, and µ s. The acceleration of gravity is g. pos. of eq. µ k, µ s ˆx (a) (5 points) A block of mass m sits in a position of equilibrium atop a spring as shown in the figure above. With the cart at rest find the constant of the spring k if the spring expands a distance d when the block is removed. Drawing the forces and using Newton s second law we get mg = kd k = mg d (b) (7 points) The cart is now accelerated with acceleration a x in the ˆx direction. Find the maximum vertical distance h that somebody inside the cart can push the weight further down from the position of equilibrium for which the static friction force will not allow the weight to come back up. Use the expression for k that you found in the previous point, and express the answer in units of d, µ s, a x, g. When the block is pushed down a further distance h, we have the forces of N, F s, f s, W. In a coordinate system with ŷ going upwards we get: ˆx :N = ma x ŷ :k(d + h) mg f s = 0, the maximum static friction force is f s = µ s N and using k = mg we get d mg d (d + h) mg µ sma x = 0 mg d h = m(g + µ sa x ) mg g d h = µ sa h = dµ sa x g Page 3

5 (c) (7 points) While the cart is still accelerating somebody pushes the block down a distance h from the position of equilibrium and then releases it to see the spring bringing the block back up. Find the acceleration of the block in the vertical direction an instant after it is released. When the block is pushed down a further distance h, we have the forces of N, F s, f k, W. In a coordinate system with ŷ going upwards we get: ˆx :N = ma x ŷ :k(d + h ) mg f k = ma y the kinetic friction force is f k = µ k N and using k = mg we get d mg d (d + h ) mg µ k ma x = ma y ma y = mg d h mµ k a x a y = g h d µ ka x (d) (7 points) Using the equation of conservation of energy compute the velocity of the block when is returning back and passing by its position of equilibrium after being released from rest from a distance h below its position of equilibrium.use the expression for k that you found in the first point. When the block is pushed down a further distance h, we have the forces of N, F s, f k, W. Let s make sure to include all the forces one way or another into the equation of conservation of energy. Putting the zero of the gravitational potential energy at the compressed distance we get: K }{{} I +U i,spring + U I,grav +W Other = K f + U f,spring + U f,grav }{{} =0 =0 1 1 k(h + d) + W Other = 1 mv f + 1 kd + mgh mg d (h + d) µ k ma x h = 1 mv f + 1 mg d d + mgh g d (h + d) µ k a x h gh = v f + g d d g( h d + h + d) µ k a x h gh = v f + gd v f = g h d µ ka x h It makes sense: v f goes to zero as h goes to zero. Page 4

6 Part 3: Funnel (4 points) A block of mass m is moving in uniform circular motion of radius inside a funnel of aperture angle α at speed v. The block is tied up to a rope that goes down the funnel as shown in the picture. An unknown force F is being applied on the rope. (a) (8 points) Find the force F on the rope that keeps the block in uniform circular motion. α F ẑ : Nsin(α) mg F cos(α) = 0 N = ˆr : Ncos(α) F sin(α) = m v from which we get (F cos(α) + mg) cos(α) + F sin(α) = mv sin(α) [ ] cos(α) F sin(α) + sin(α) = m v cos(α) mg sin(α) F = m v sin(α) mg cos(α) F cos(α) + mg sin(α) (b) (8 points) If you were to re-design the funnel such that when you have the block in the same conditions the force is zero, what aperture angle would you choose? From the previous point we got: F = m v sin(α) mg cos(α) so choosing α such that F = 0 gives us m v sin(α) = mg cos(α) tan(α) = g v Page 5

7 (c) (8 points) If the rope now is shortened such that the block is now rotating at with speed v. What was the work done in shortening the rope? Using conservation of enery we get: K i + U i + W other = K f + U f 1 mv + mg tan(α) + W other = 1 m(v) + mg tan(α) W other = 4 1 mv 1 mv + mg tan(α) mg tan(α) W other = 3 1 mv mg tan(α) Page 6

8 Part 4: Pulleys (30 points) A system of two pulleys is constructed as shown in the diagram below. The string and both pulleys are masless and the masses of the blocks are known and such that m 1 = 3m. The gravity constant is g. pulley# pulley#1 m m 1 (a) (8 points) Draw the free-body diagrams of the each pulley and the masses. T T T 3 T T pulley#1 m 1 pulley# m T 3 m 1 g T T m g (b) (6 points) Put a coordinate system. If mass m moves up a distance Y, what is the change in vertical position of m 1? Taking derivatives reach a relationship between the acceleration of both blocks a 1y, and a y. Y 1 = Y a 1Y = a Y Page 7

9 (c) (8 points) Compute the tension in the string and the acceleration of mass m 1. Express the tension in terms of m. From the free-body diagrams we have : pulley #1 : T T 3 = 0 m 1 : T 3 m 1 g = m 1 a 1y pulley # : T T = 0 m : T m g = m a y = 0 From the two first we get: T m 1 g = m 1 a 1Y From the 4th eq we get: T m g = m a Y = m a 1y T = m g m a 1y, which replacing this T in to the above one: m g 4m a 1 y m 1 g = m 1 a 1Y g(m m 1 ) = a 1y (m 1 + 4m ) a 1y = g m m 1 = g m 1 + 4m 7 T = m g m a 1 y = m g + m g 7 = T = m g 9 7 (d) (8 points) Compute the tension in the string connecting pulley # to the ceiling. Express in terms of m From the free-body diagrams we have : ŷ : T T = 0 T = T = m g 18 7 Page 8

10 Part 5: Skiing (16 points) A 75 kg ski-jumper starts from rest at top of a 10m long ramp which makes an angle of α = 37 to the horizontal. There is friction between her skis and the ramp. Ignore the friction due to the air. α = 37 (a) (8 points) If her speed at the bottom is v=5 m s that retarded her descent? what is the frictional force I call d=ramp length=10 m. Using the conservation of energy equation: mgdsin(α) f k d = 1 mv f K = mgsin(α) 1 d mv f k = 47N (b) (8 points) What is the coefficient of kinetic friction? Solution µ k = f k N = µ k = N mg cos(37 ) Page 9