Physics 2210 Fall Review for Midterm Exam 2 10/07/2015

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1 Physics 2210 Fall 2015 Review for Midterm Exam 2 10/07/2015

Problem 1 (1/3) A spring of force constant k = 800 N/m and a relaxed length L 0 = 1.10 m has its upper end fixed/attached to a pivot in the ceiling. Its free end is attached to a block of mass m=45.

2 Problem 1 (1/3) A spring of force constant k = 800 N/m and a relaxed length L 0 = 1.10 m has its upper end fixed/attached to a pivot in the ceiling. Its free end is attached to a block of mass m=45.0 kg that sits on a frictionless, horizontal surface. When the block is directly below the pivot, the spring is stretched to a length h = 1.20 m. The block is then pulled left a horizontal distance b = 0.90 m and released from rest. Calculate (a) the magnitude of the force exerted by the spring on the block, (b) the acceleration of block right after being released, and (c) the speed of block as it passes directly under the pivot again. You can assume that the block remains on the surface. Solution: (a) F s = k L i, L i =L i -L 0, L 0 =1.10m, k=800n/m L i = h 2 + b 2 = 1.20m m 2 = 1.50m F s = 800N/m 1.50m 1.10m = 320N

Problem 1 (2/3) Spring: k = 800 N/m, L 0 = 1.10 m attached to a block of mass m=45.0 kg on a frictionless, horizontal surface.

(c) the speed of block as it passes directly under the pivot again Solution: (b), we had F s =320N If the block stays on the surface then

3 Problem 1 (2/3) Spring: k = 800 N/m, L 0 = 1.10 m attached to a block of mass m=45.0 kg on a frictionless, horizontal surface. Directly below pivot, the spring is stretched to h = 1.20 m. The block is then pulled left a horizontal distance b = 0.90 m and released from rest. Calculate (a) the magnitude of the force exerted by the spring on the block, (b) the acceleration of block right after being released, and (c) the speed of block as it passes directly under the pivot again Solution: (b), we had F s =320N If the block stays on the surface then its acceleration is purely to the right, due to the x-component of the net force on it. a y = 0, and F x = F s cos φ = ma x, φ = tan 1 h b = tan m 0.90m = tan = a x = F s sin φ 320N cos = = 4.27 m s 2 m 45kg a = 4.27 m s 2 i or 4.27 m s 2 to the right

4 Problem 1 (3/3) Spring: k = 800 N/m, L 0 = 1.10 m attached to a block of mass m=45.0 kg on a frictionless, horizontal surface. Directly below pivot, the spring is stretched to h = 1.20 m. The block is then pulled left a horizontal distance b = 0.90 m and released from rest. Calculate (c) the speed of block as it passes directly under the pivot again Solution (c) Normal force does no work (perpendicular to the path), and there is no friction. There is no change in height gravity does no work. Only the spring force is involved and it is a conservative force. Total energy is conserved: E i = K i + U i = 1 mv 2 i k ΔL 2 i 2 = k ΔL 2 i 2 (initially at rest:v i =0) E f = K f + U f = 1 mv 2 f k ΔL 2 f 2, and Ef = E i 1 2 f k ΔL 2 f 2 = 1 2 i 2 mv 2 f = k ΔL 2 2 i ΔL f From part (a) ΔL i =0.40m and ΔL f = h L 0 = 1.20m 1.10m = 0.10m v 2 f = 800 N m 0.40m m kg = m 2 s 2 v f = 1.63 m s

5 Problem 2 (1/3) A block in the figure has mass m 1 = 6.10 kg. The coefficient of static friction between the block and the horizontal desk top is μ s = The block is attached by a cord to a cowbell of mass m 2 = 1.25 kg. The cowbell is attached by a second cord to a hook on the wall. The first cord is horizontal, and the second cord at an angle of θ = 40.0 from the horizontal. (a) The system is in equilibrium as shown. Calculate the tension T 2 on the block. (b) Calculate the magnitude of the friction force exerted by the desk on the block. (c) How much additional mass can be added to the cowbell (i.e. to m 2) and still keep the system in equilibrium? Solution: (a) consider the freebody diagram of the cowbell. The system is at equilibrium acceleration of the cowbell is zero. We start with the y-component of the net force on the cowbell: F 2y = T 2 sin θ m 2 g = m 2 a 2y = 0 T 2 sin θ = m 2 g T 2 = m 2g 1.25 kg 9.8 m s2 = = 19.1 N sin θ sin 40.0

Problem 2 (1/3) m 1 = 6.10 kg. μ s = 0.40. m 2 = 1.25 kg. θ = 40.0 from the horizontal. (a) The system is in equilibrium as shown. Calculate the tension T 2 on the block.

Solution: (b) Looking again at the free-body diagram: The x-component of the net force on the cowbell: F 2x = T 2 cos θ T 1 = m 2 a 2x = 0 T 1 = T 2 cos θ T 1 = m 2g sin θ cos θ = m 2g cot θ Now on

6 Problem 2 (1/3) m 1 = 6.10 kg. μ s = m 2 = 1.25 kg. θ = 40.0 from the horizontal. (a) The system is in equilibrium as shown. Calculate the tension T 2 on the block. (b) Calculate the magnitude of the friction force exerted by the desk on the block. (c) How much additional mass can be added to the cowbell (i.e. to m 2) and still keep the system in equilibrium? Solution: (b) Looking again at the free-body diagram: The x-component of the net force on the cowbell: F 2x = T 2 cos θ T 1 = m 2 a 2x = 0 T 1 = T 2 cos θ T 1 = m 2g sin θ cos θ = m 2g cot θ Now on to the free-body diagram on the block: x-component of net force: F 1x = T 1 f S = m 1 a 1x = 0 f S = T 1 = m 2 g cot θ f S = 1.25 kg 9.8 m s 2 cot 40.0 = 14.6N Compare this to maximum static friction force given by f SMMM = μ s N. Now, the y-component of the net force on the block is: F 1y = N m 1 g = m 1 a 1y = 0 N = m 1 g f SSSS = μ s m 1 g = kg 9.8 m s 2 = 23.9N > 14.6N

7 Problem 2 (1/3) m 1 = 6.10 kg. μ s = m 2 = 1.25 kg. θ = 40.0 from the horizontal. (a) The system is in equilibrium as shown. Calculate the tension T 2 on the block. (b) Calculate the magnitude of the friction force exerted by the desk on the block. (c) How much additional mass can be added to the cowbell (i.e. to m 2) and still keep the system in equilibrium? Solution: (c) From part (b) we had f S = T 1 = m 2 g cot θ, and f SSSS = μ s m 1 g = 23.9N. So if we were to increase m 2, then f S increases proportionally So the maximum total m 2 corresponds to the case f S = f SSSS : f SSSS = m 2MMM g cot θ m 2MMM = f SSSS g 23.9N tan θ = 9.8 m s 2 tan 40.0 = 2.05kg And so the maximum additional mass on the cowbell that still maintains equilibrium is m 2MMM = m 2MMM m 2 = 2.05kg 1.25kg = 0.80 kg

8 Problem 3 (1/3) In the figure, the pulley has negligible mass and is frictionless. Block A has a mass of 2.5 kg, and block B 4.0 kg. The angle of the incline is θ = 35. The coefficient of kinetic friction between the incline and block A is μ k = The blocks are set in motion from rest with the cord taut. (a) Find the magnitude of the friction force exerted by the incline on block A. After block B falls a distance of D=45 cm, calculate N T (b) the work done by gravity on block A, and (c) the total kinetic energy of the two blocks. f k You can assume that block A remains on the incline. m A g Solution: (a) Since m B > m A block B will clearly drop and A will move up the incline. So the kinetic friction force acts down the incline. For analyzing block A, we choose +x to be up the incline, and +y perpendicularly out of the incline. Y-component of net force on A: F AA = N m A g cos θ = m A a AA = 0 N = m A g cos θ f k = μ k N = μ k m A g cos θ = kg 9.8 m s 2 cos 35 = 6.02N

9 Problem 3 (2/3) Pulley is massless and frictionless. m A =2.5 kg, m B =4.0 kg. θ = 35. μ k = 0.30 between the incline and block A. Blocks are set in motion from rest (a) Find the magnitude of the friction force exerted by the incline on block A. After block B falls a distance of D=45 cm, calculate (b) the work done by gravity on block A, and (c) the total kinetic energy of the two blocks. m A g Solution: (b) W gg = F g r A = m A gd cos θ + 90 = 2.5kg 9.8 m s m cos 125 Note: cos a + b = cos a cos b sin a sin b cos θ + 90 = cos θ cos 90 sin θ sin 90 = sin θ And we actually have cos 125 = sin35 = W gg = 2.5kg 9.8 m s m = 6.32J Alternately: By the definition of potential energy: D θ W gg = U gg, and U gg = m A g h A. Here block A moved a distance D up an incline and so h A = D sin θ W gg = m A g h A = m A gd sin θ which is the same answer we had before. h A

10 Problem 3 (3/3) Pulley is massless and frictionless. m A =2.5 kg, m B =4.0 kg. θ = 35. μ k = 0.30 between the incline and block A. Blocks are set in motion from rest (a) Find the magnitude of the friction force exerted by the incline on block A. After block B falls a distance of D=45 cm, calculate (b) the work done by gravity on block A, and (c) the total kinetic energy of the two blocks. Solution: (c) We will consider block A + block B as a single object. Since we have friction acting on A (the only non-conservative EXTERNAL force), then we have (unit 09): E = W k = f k r A = f k D = 6.02N 0.45m = 2.71J (by the way, the force of the cord is internal to the system) Initially, the blocks are at rest: K i = 0, we are asked to find K f = K i + K = K But by the definition of total energy: E = + U, and U = U gg + U gb U gg = m B g h B = m B gd = 4.0kg 9.8 m s m = 17.64J And so K f + U gg + U gg = W k K f = U gg U gg + W k = 6.32J J 2.71J = 8.61J

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