Physics 110 Homework Solutions Week #5

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1 Physics 110 Homework Solutions Week #5 Wednesday, October 7, 009 Chapter C 5. A 5.8 B A crate on a ramp a) x F N 15 F 30 o mg Along the x-axis we that F net = ma = Fcos15 mgsin30 = 500 cos15 75(9.8)sin30 = 115 N; then a = F net /m = 115/75 = 1.54 m/s up the plane b) v = v o + ax gives us v = 0 + (1.54)5 or v = 3.9 m/s c) W F = Fcos15 x = 500(5)cos15 = 415 J d) W grav = -mgh = -mgdsin30 = -75(9.8)(5)sin30 = J e) First note that W FN = 0; then W net = ΔKE = = 575 J; so ½ mv 0 = 575 J so that v = 3.9 m/s 5.35 A toy car a) E init = mg H init = mg(1. m); E top = mg H top + ½ mv top Conservation of energy then implies that: mg(1.) = mg(0.5) + ½ mv top Or v top = [(1.)g 0.5g] = 1.9 g or v top = 4.3 m/s b) At the bottom E b = 0 + ½ mv b ; then conservation of energy gives mg(1.) = ½ m v b ; or v b =.4 g so that v b = 4.85 m/s both on way up and down c) E init = mgh with H to be determined. Conservation of energy gives us mgh = E top = mg(0.5) + ½ mv We two unknowns H and v; but at the top a free body diagram would only two downward forces (mg and F N (why is F N down here?) and to just barely make it to the top, the car will be barely in contact with the track and so F N = 0 and the only force acting is mg. The centripetal force mg is then equal to ma = mg and so v /r = g or v = rg = (0.5/)g. Then we can write mgh = 0.5mg + ½ m(0.5/)g = 0.31 mg and so H = 0.31 m The initial energy stored in the spring as potential energy is converted to a gravitational potential energy as the marble rises vertically and travels a distance D

2 along the ramp. To determine the distance D we use conservation of energy and we ΔU s + ΔU g = 1 kx f 1 kx i ( ) + ( mgy f mgy i ) = 1 kxi + mgy f = 0 ( ) kx i mgsinθ = 10 N 0.03m m y f = Dsinθ = kx i mg D = = 0.175m =17.5cm. 0.05kg 9.8 m sin3 s Since this is less than the 60cm the ball does not make it. In order to the ball travel 60cm we need to compress the spring by an amount given by ΔU s + ΔU g = 1 kx f 1 kx i x i = ( ) + ( mgy f mgy i ) = 1 kxi + mgy f = 0 mgdsinθ = 0.05kg 9.8 m 06msin3 s k 10 N m = 0.055m = 5.5cm Thursday, October 8, 009 Chapter B 5.37 A roller coaster a. Applying conservation of energy between the top of the crest, point #1 and point #, 10 m up the second hill we ΔKE + ΔU g = 1 mv f, 1 mv i,1 v f, ( ) + ( mgy f, mgy i,1 ) = 0 = g( y i,1 y f, ) + v i,1 v f, = 9.8 m 10m + ( 10 m s s ) =17. m s b. The maximum height of the second hill is found when the velocity of the cart goes to zero. Applying conservation of energy between point #1 and this maximum height, point # we ΔKE + ΔU g = 1 mv f, 1 mv i,1 ( ) + ( mgy f, mgy i,1 ) = 0 y f, = y i,1 + v i,1 g y = 0m + f, ( 10 m s ) = 5.1m 9.8 m s c. Here a frictional force causes the cart to lose 8000J of energy. The maximum height is found from ΔKE + ΔU g = 1 mv f, 1 mv i,1 ( ) + ( mgy f, mgy i,1 ) = ΔE = 8000J y f, = y i,1 + v i,1 g 8000J mg y f, m ( 10 s ) = 0m m s 8000J 500kg 9.8 m s = 3.5m

3 5.39 A block on a ramp connected to a hanging mass a. First we need free-body diagrams for each block: F N T T m 1 g m g Then we write Newton s second law equations for each block: T m 1 gsin30 = m 1 a and m g T = m a T and a are unknowns; eliminate T (easiest to add both equations together): m g m 1 gsin30 = (m 1 + m )a and solving for a we a = (m g m 1 gsin30)/(m 1 + m ) = 4.15 m/s b. If a = 4.15 m/s, x = m, v o = 0 then v = ax gives v = [(4.15)()] = 4.07 m/s 5.48 Applying conservation of energy we ΔKE + ΔU = mgh mgh = F fr x = µ k mgx x = H h µ k 5.71 An amusement park thrill ride a. Applying conservation of energy we b. To determine the speed of the cart before the loop-the-loop we apply conservation of energy between the top of the ramp and the bottom. Thus we c. Again apply conservation of energy between the bottom and the top of the loopthe-loop and we d. The work done is the difference in the kinetic energies and we e. To calculate the new speed we consider the energy dissipated by friction. Noting that the difference in energy is the energy dissipated as friction we

4 Friday, October 9, 009 Chapter C 5.4 D 5.5 D 5.36 A loop-the-loop roller coaster a) E initial = mgh, with H = 15 m ; E final = mgr + ½ mv, where R = radius of loopthe-loop; Using conservation of energy, we mgh = mgr + ½ mv ; solving, we find v = 15.0 m/s b) When at ground level, the energy is all KE and using conservation of energy we ½ mv = mgh, or v = sqrt(gh) = 17.1 m/s c) When at the position in part (a) the forces acting are the weight (vertically down) and the normal force (horizontally directed toward the center of the loop). The normal force is then the net centripetal force and must equal mv /R = (500)(15) /3.5 = 3140 N; while the weight is mg = (500)(9.8) = 4900 N. The net force is then equal to (by Pythagorean theorem) sqrt((4900) + (3140) ) = 3500 N directed at an angle of θ = tan -1 (4900/3140) = 8.6 o below the horizontal 5.38 Applying conservation of energy between the release point of the block and when the spring is compressed we, with ΔKE = 0 since the block starts at rest and returns to rest when the spring is fully compressed. Thus. Choosing x i (the spring is uncompressed) and y f (the zero of the gravitational potential energy) equal to zero we, where the height the block fell through is given by the geometry of the system as y i = dsinθ A box on an incline

5 a. Choosing down the incline as the positive direction for the horizontal forces we the net work given as, since the displacement is opposite to the force of gravity on the way up and in the same direction on the way down. b. The frictional force opposes the motion on the way up and on the way down. The net work is therefore. The distance traveled by the block is determined from the time independent equation of motion, where the final velocity of the block is zero and the acceleration is determined from Newton s nd law. We therefore c. The net change in the energy during the round trip is the sum of the work done by the force of gravity and by the frictional force. Thus the net work done is 1.03J Monday, October 1, 009 Chapter C 6.7 D 6.8 B 6.9 C 6.1 It s velocity as the ball hits the ground is found from v = v o + ay with v o = 0m/s, a = 9.8m/s and y = 1 m; so v = 4.43 m/s down; then p = mv =.1 kg m/s down A ball bouncing off of a wall a. Change in momentum = p final p init (both are vectors in bold); but the direction reverses after the collision and so Δp = 0.1(5) (-0.1)(5) = 1 kgm/s away from the wall b. Since the average force times the collision time = impulse = change in momentum, we that the average force = (change in momentum)/(collision time) = 1/.005 s = 00 N away from the wall c. Yes it did, because the momentum of the (ball + wall) is conserved. The force on the wall from the ball is equal and opposite of the force on the ball from the wall

6 and so Δp wall + Δp ball = 0 and Δp wall = -Δp ball ; but the M wall is so large that the v wall is negligible miles per hour translates to 55.6m/s. For a tennis ball launched from rest, the change in the momentum of the object is given as. By the impulse momentum theorem, (Newton s second law) the average force is A railroad car a. Using conservation of momentum, the final velocity is given by 10,000kg(4m/s) + 0 = (10,000kg+1,00kg)v final or v final = 1.4 m/s in the direction the railroad car was traveling. b. KE init = ½ (10,000)(4) =.88 x 10 6 J KE final = ½ (11,00)(1.4) =.56 x 10 6 J, so the % loss is [KE init KE final ]/KE init x 100 = 11.1% c. Frictional force = µ k F N = 0.9(11,00)(9.8) = 9.88 x 10 4 N. d. Work by friction = ΔKE = 0 ½ mv = -1/ (11,00)(1.4) = -.56 x 10 6 J A roller coaster a. Using conservation of energy between the initial point and point A we the speed of the object as where we take the zero of the gravitational potential energy to be at ground level. The centripetal force, directed vertically upward at point A, has magnitude. b. To determine the speed at point B we use conservation of energy between points A and B. We,. c. Point C is at the zero of gravitational potential energy and given that energy is conserved, the speed of the car at point C has to be the same as at point A or 14 m/s. Using conservation of momentum we the right. directed to