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Extra exercises to lectures 3 and 4 Exercise 7.20 A small particle of mass m is pulled to the top of a frictionless half-cylinder (of radius R ) by a cord that passes over the top of the cylinder, as indicated in the figure. a- if the particle moves at constant speed, show that F = mgcos b- by directly integrating W F.dr find W when the particle is moving from the bottom to the top. Can you explain your results? F Solution 7.20 x F n a)- The radius to the object makes angle with the horizontal, so its weight makes the same R angle with the negative side of the x-axis, when we take the x-axis in the direction of motion tangent to the cylinder. mg F x max 0 F mg cos 0 F mg cos b)- We use radian measure to express the next bit of displacement as Rd f f f f W F .d Fd FRd R Fd i i i i /2 R mg cos d mgR[sin( / 2) sin(0)] mgR 0 This is the same if we calculate the difference in potential energy between the bottom and the top, and the result is independent of the pass, thus along the quarter of the circle or along the blue pass we get the same results. The weight is a conservative force. Exercise 7.32 A 2.00kg block is attached to a spring of force constant 500N/m, along the x- direction. The block is pulled xm= 5cm to the right of equilibrium and released from rest. Find the speed of the block as it passes through equilibrium if: a- the horizontal surface is frictionless, b- the coefficient of friction between the block and surface is 0.350 xm 0 Equilibrium Solution 7.32 a)- The conservation of total mechanical energy in absence of friction gives K U 0 1 2 1 2 1 2 1 2 Ki U i K f U f mvi kxi mv f kx f 2 2 2 2 1 1 2 k xi xm ; x f xequil 0; vi 0; mv 2 kxm v f f xm =0.791 2 2 m m/s b)- The presence of friction destroy the conservation of energy K U f .x ; f k N k mg 1 2 1 2 1 2 1 2 K i U i K f U f f .x mvi kxi mv f kx f f .x 2 2 2 2 1 2 1 2 xi xm ; x f xequil 0; vi 0; mv f kxm k mgxm 2 2 k 2 k 2 v 2 xm 2 k gxm v f f xm 2 k gxm 0.531m / s m m Exercise 8.23 A force acting on a particle moving in the xy plane is given by F (2 yx x 2 y ) N ; ˆ ˆ x, y in meters The particle moves from the origin to a final position having coordinates x = 5.00m and y y = 5.00m. Calculate the work done by F along the three paths indicated. Explain your results. B C Solution 8.23 a)- along OAC x 5 5 O A WOA dxx.(2 yx x 2 y ) 2 y y 0 dx 0 ˆ ˆ ˆ 0 0 5 5 WAC dyy.(2 yx x y ) x 2 ˆ ˆ ˆ 2 dy 125 J 0 0 x 5 WOAC WOA WAC 0 125 J 125 J b)- along OBC 5 5 WOB dyy.(2 yx x y ) x 2 ˆ ˆ ˆ 2 dy 0 0 0 x 0 5 5 WBC dxx.(2 yx x y ) 2 y y 5 dx 50 J ˆ ˆ ˆ 2 0 0 WOBC WOB WBC 0 50 J 50 J c)- along OC 5 5 WOC (dxx dyy ).(2 yx x y ) (2 ydx x 2 dy ) ˆ ˆ ˆ ˆ 2 0 0 5 along OC, x y dy dx WOC (2 xdx x 2 dx) 0 WOC 66.7 J d)- the force F is a non-conservative force since the work is path dependent. Exercise 8.33 A 5.00kg block is set into motion up inclined plane with an initial speed of 8.00 m/s. The block comes to rest after traveling 3.00m along the plane, which is inclined at an angle of 30o to the horizontal. For this motion determine: 1. The change in the block’s kinetic energy, 2. The change in the potential energy of the block-Earth system, 3. The friction force exerted on the block ( assumed to be constant) 4. What is the coefficient of kinetic friction? 8m/s 3m 30o Solution 8.33 1 1 a) The change in kinetic energy is K m(v 2 vi2 ) mvi2 160 J f 2 2 b) The change in potential energy is U mg (3m) sin 30o 73.5 J c) The mechanical energy converted due to friction is K+U= - f x= 86.5 J Hence f =86.5/3 = 28.8N d) f= k n =k mgcos30o = 28.8 J Hence, the coefficient of friction is k =0.679N Exercise 8.48 A block slides down a curved frictionless track and then up an inclined plane as in the figure. The coefficient of kinetic friction between the block and incline is k = 0.325. Use energy methods to show that the maximum height reached by the block is h ymax 5.1m 1 k cot h = 8m ymax 30o Solution 8.48 The potential energy of the block-Earth system is mgh The amount of energy k mgd cos is converted into internal energy due to friction on the incline. Therefore the final height ymax is found from K U fd 0 U f U i fd ymax mgymax mgh k mgd cos ; d sin Solving h ymax 1 k cot Using the data we get the answer given Exercise 8.70 A ball is tied to one end of a string. The other end of the string is held fixed. The ball is set moving around a vertical circle without friction, and with speed v Rg at the top of the circle, as it is shown in the figure. At what angle should the string be cut so that the ball will then travel through the center of the circle? Solution 8.70 vo v Rg At the top position, if the tension is zero then v2 R mg m v Rg C R At any angle , the conservation of energy gives us the speed vo . Take the origin of potential energy at the bottom position , ( =- ) Ki U i K f U f 1 2 1 2 mvi mg 2 R vo mg ( R R cos ) vo (3 2cos ) gR 2 2 What we need is now to solve the projectile problem with initial velocity components vxi vo cos ; v yi vo sin The requirement is that the trajectory ( parabola ) have to pass through the center C. the origin is taken the position of the point when we cut the string 1 g y x 2 tan x yo 2 vo cos 2 2 0 The center has coordinates in the x,y axis with origin at the position of the mass with speed vo : y xC R sin ; yC R cos x O Substituting and using sin 2 cos 2 1 R C Simplifying, to get 6 36 12 3cos 6 cos 1 0 cos 2 6 The only physical solution is with the negative sign . Thus cos 0.1835 79.43o 100.6o Exercise 8.72 A pendulum, comprising a string of length L and a small sphere, swings in the vertical plane. The string hits a peg located a distance d below the point of suspension. a- show that if the sphere is released from a height below that of the peg, it will return to this height after striking the peg. b- show that if the pendulum is released from the horizontal position ( = 90o ) and is to swing in a complete circle centered on the peg, then the minimum value of d must be 3L/5. d L peg Solution 8.72 a) Energy is conserved in the swing of the pendulum, and the stationary peg does no work. So the ball’s speed doesn’t change when the string hits the peg, and the ball swings equally high on both sides. b) Relative to the point of suspension, U i 0, U f mg[d ( L d )] From this we find 1 2 mg (2d L) mvi 0 2 Also for centripetal motion vi2 mg m where R L d vi2 Rg ( L d ) g R Solve these equations to get L d 3 2d L 0 d L 2 2 5 Exercise 9.4 Two blocks of masses M and 3M are placed on a horizontal, frictionless surface. A light spring is attached to one of them , and the blocks are pushed together with the spring between them. A cord initially holding the blocks together is burned; after this, the blocks of masse 3M moves to the right with a speed 2.00m/s. a- What is the speed of the block of mass M? b- Find the original elastic potential energy in the spring if M = 0.35 kg Solution 9.4 For the system of two blocks p 0 Mvm 3m 2 0 vm 6.00 Motion to the left b) 1 2 1 1 kx MvM 3Mv3M 8.40 J 2 2 2 2 2 Exercise 9.12 A professional diver performs a dive from a platform 10m above the water surface. Estimate the order of magnitude of the average impact force she experiences in her collision with the water. The impact time take it 1.0s and the mass of the diver is 55 kg. Solution 9.12 If the diver starts from rest and drops vertically into the water, the velocity just before the impact is found from 1 2 Ki U i K f U f vimpact 0 0 mgh vimpact 2 gh 2 With the diver at rest after an impact time t , the average force during impact is given by the second Newton’s law m(0 vimpact ) m 2 gh Fav t t Using the given numbers: Fav 770N Exercise 9.24 As shown in the figure, a bullet of mass m and speed v passes completely through a pendulum bob of mass M. The bullet emerges with a speed of v/2. The pendulum bob is suspended by a stiff rod of length l and negligible mass. What is the minimum value of v such that the pendulum bob will barely swing through a complete vertical circle? l m M v v/2 Solution 9.24 Energy is conserved for the bob-Earth system between bottom and top of swing. At the top the stiff rod is in compression and the bob nearly at rest 1 Ki U i K f U f Mvb 0 0 Mg 2 2 2 vb g 4 vb 2 gf 2 Momentum of the bob-bullet system is conserved in the collision v 4M mv m M (2 g ) v g 2 m Exercise 9.27 a) Three carts of masses 4 kg, 10 kg and 3 kg move on a frictionless horizontal track with speeds of +5m/s, +3m/s, and - 4m/s respectively. The carts stick together after collision. Find the final velocity of the train of three carts. b) Does your answer require that all the carts collide and stick together at the same time? What if they collide in a different order? 4m/s 3m/s -4m/s Solution 9.27 a)- Using conservation of momentum for a perfect inelastic collision: p before p after m1v1 m2v2 m3v3 (m1 m2 m3 )vG v 2.24m / s To the right b)- No. For example, if m2 and m3 were to stick first, they would move with a speed m2 v2 m3v3 (m2 m3 )vG 23 Then, when this 13 kg =(m2 + m3) collides with the 4 kg , we have m1v1 (m2 m3 )vG 23 m1v1 m2v2 m3v3 (m1 m2 m3 )vG Thus coupling order makes no difference Exercise 9.58 A bullet of masse m is fired into a block vi M m of mass M initially at rest at the edge of a frictionless table of height h. The bullet remains in the block, and h after impact the blocks lands a distance d from the bottom of the table. Determine the initial speed of the bullet. d Solution 9.58 M m mvi ( M m)v f v f vi m 1 2 2h d g gd 2 d v f t ; h gt t g and v f d 2 t 2h 2h M m gd 2 This gives vi m 2h

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