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Physics 2210 Homework 8 Spring 2015 Charles Jui February 23, 2015 IE Block Spring Incline Wording A 5 kg block is placed near the top of a frictionless ramp, which makes an angle of 30◦ degrees to the horizontal. A distance d = 1.3 m away from the block is an unstretched spring with k = 3 × 103 N/m. The block slides down the ramp and compresses the spring. Find the maximum compression of the spring. See Figure 1. Figure 1: IE Block Spring Incline 1 Solution Since only conservative forces are at play, the total mechanical energy is conserved 1 , i.e. M Ei = M Ef mg(xmax + d) sin θ = 1 2 kx 2 max which yields quadratic equation in xmax x2max − 2mgd 2mg xmax − =0 k k and its solutions are x± max mg sin θ = k √ ( 1± 2kd 1+ mg sin θ ) We eliminate the negative solution since it would mean that the block is not in contact with the spring. And we will get xmax = 0.1541949 m Tipler6 7.P.012. Wording Estimate the following: (a) The change in your gravitational potential energy on taking an elevator from the ground floor to the top of the Empire State Building. The building is 102 stories high (assuming a 3 min ride to the top of the building). (Assuming your mass is 68 kg and the height of one story to be 3.5 m.) (b) The average force exerted by the elevator on you during the trip. (c) The average power delivered by that force. Solution (a) Change in potential energy is given by ∆U = mg∆h where ∆h = 102 × 3.5 m = 357 m, thus ∆U = 238.14756 kJ 1 Note adding xmax in the potential energy. 2 (b) The average force is defined by F̄ = ∫ 1 h2 − h1 h2 F · dx h1 but ∆h = h2 − h1 and the integral is simply the work done, which in this case is equal to the change in potential energy, i.e. F̄ = ∆U ∆h So we get F̄ = 667.08 N (c) The average power is defined as P̄ = W ork ∆U = time ∆t where the time is t = 3.5 × 60 s = 210 s, so P̄ = 1.323042 kW Tipler6 7.P.034. Wording A simple Atwood’s machine uses two masses, m1 and m2 . Starting from rest, the speed of the two masses is 3 m/s at the end of 7 s. At that instant, the kinetic energy of the system is 54 J and each mass has moved a distance of 10.5 m. Determine the values of m1 and m2 . See Figure 2. Solution Here we have conservation of energy: At the beginning, the total energy is zero.2 After a while 0= 1 1 m1 v12 + m2 v22 + m1 gh1 + m2 gh2 2 2 Let’s assume that the first mass goes down; now we have h2 = −h1 = d Hence we obtain 0= 2 We v1 = −v2 = v 1 (m1 + m2 )v 2 + (m2 − m1 )gd 2 can pick the masses to be at the same height, for convenience. 3 Figure 2: Tipler6 7.P.034. We are also given the total kinetic energy, i,e, KE = 1 1 1 m1 v12 + m2 v22 = (m1 + m2 )v 2 2 2 2 Combining the two equations produces the following system = (m1 + m2 )v 2 = (m1 − m2 )gd 2KE KE Solving this system yields (a) ( m1 = 1 1 + v2 2gd ) KE m1 = 6.262 kg (b) ( m2 = 1 1 − v2 2gd ) KE m2 = 5.738 kg Tipler6 7.P.036. Wording A block of mass m is pushed up against a spring, compressing it a distance x, and the block is then released. The spring projects the block along a frictionless horizontal surface, giving the block a speed v. The same spring projects a 4 second block of mass 3m, giving it a speed of 4v. What distance was the spring compressed in the second case? Give your answer in terms of a numerical factor times x, the compression of the first block. Solution Transfer of energy in the first case 1 2 1 kx = m1 v12 2 2 Transfer of energy in the second case 1 1 2 ky = m2 v22 2 2 Taking ratio of the two equations yields y2 m2 v22 m2 v22 = = · x2 m1 v12 m1 v12 √ Thus y= √ 3m 4v x = 4 3x m v y = 6.928 x Tipler6 7.P.060. Wording In a volcanic eruption, 2km3 of mountain with a density of 1600 kg/m3 was lifted an average height of 513 m. (a) How much energy in joules was released in this eruption? (b) The energy released by thermonuclear bombs is measured in megatons of TNT, where 1 megaton of TNT = 4.2 × 1015 J. Convert your answer for (a) to megatons of TNT. Solution (a) The energy required is equal to the change of potential energy of the ash, i.e. E = ∆U = M g∆h = ρV g∆h where V = 2 × 109 m3 . This gives value E = 1.610E + 16 J 5 (b) Simple ratio E/(4.2 × 1015 J) yields 3.834 Mton TNT Pendulum Wording A mass m = 4.7 kg hangs on the end of a massless rope L = 2.05 m long. The pendulum is held horizontal and released from rest. 1) How fast is the mass moving at the bottom of its path? 2) What is the magnitude of the tension in the string at the bottom of the path? 3) If the maximum tension the string can take without breaking is Tmax = 382 N , what is the maximum mass that can be used? (Assuming that the mass is still released from the horizontal and swings down to its lowest point.) 4) Now a peg is placed 4/5 of the way down the pendulum’s path so that when the mass falls to its vertical position it hits and wraps around the peg. As it wraps around the peg and attains its maximum height it ends a distance of 3/5 L below its starting point (or 2/5 L from its lowest point). How fast is the mass moving at the top of its new path (directly above the peg)? 5) Using the original mass of m = 4.7 kg, what is the magnitude of the tension in the string at the top of the new path (directly above the peg)? Figure 3: Pendulum 1 6 Figure 4: Pendulum 2 Solution 1) Forces acting on the pendulum are conservative, thus the energy is conserved, i.e. 1 mgL = mv 2 2 so we get √ v = 2gL and numerically v = 6.342 m/s 2) At the bottom, the vertical component of the acceleration must be equal to centripetal acceleration. At the same time, the net force is made up of gravity and tension; hence according to the II.Newton law T − mg = m which gives3 v2 L ( ) ( ) v2 2gL T =m g+ =m g+ = 3mg L L and thus T = 138.321 N 3) Turning around the result of the previous part mmax = Tmax 3g mmax = 12.980 kg 3 Using results of the previous part for the speed at the bottom. 7 4) Once again, all the force involved are conservative, thus the energy is conserved 1 mghi = mghf + mv ′2 2 so v′ = √ √ √ 2g(hi − hf ) = 2g(3/5)L = (6/5)gL which gives v ′ = 4.912 m/s 5) Here the approach is analogous to the one in part 3) with the modification of the net force: v ′2 T ′ + mg = m (1/5)L ( 2 ) 5v‘ ′ T =m − g = 5mg L with the value of T ′ = 230.535 N Loop the Loop Wording A mass m = 73 kg slides on a friction-less track that has a drop, followed by a loop-the-loop with radius R = 18.1 m and finally a flat straight section at the same height as the center of the loop (18.1 m off the ground). Since the mass would not make it around the loop if released from the height of the top of the loop (do you know why?) it must be released above the top of the loop-the-loop height. See the Figure 5. (Assume the mass never leaves the smooth track at any point on its path.) Figure 5: Loop the Loop 8 1) What is the minimum speed the block must have at the top of the loop to make it around the loop-the-loop without leaving the track? 2) What height above the ground must the mass begin to make it around the loop-the-loop? 3) If the mass has just enough speed to make it around the loop without leaving the track, what will its speed be at the bottom of the loop? 4) If the mass has just enough speed to make it around the loop without leaving the track, what is its speed at the final flat level (18.1 m off the ground)? 5) Now a spring with spring constant k = 15300 N/m is used on the final flat surface to stop the mass. How far does the spring compress? 6) It turns out the engineers designing the loop-the-loop didn’t really know physics – when they made the ride, the first drop was only as high as the top of the loop-the-loop. To account for the mistake, they decided to give the mass an initial velocity right at the beginning. How fast do they need to push the mass at the beginning (now at a height equal to the top of the loop-the-loop) to get the mass around the loop-the-loop without falling off the track? Solution 1) At the top of the loop, the centripetal force needs to be at least as large as the force of gravity, hence v2 Fg = m min R so we get √ vmin = gR and numerically vmin = 13.325 m/s 2) Requirement of the previous part puts the total energy of the box at the top of the loop to 1 2 + mg(2R) T E = mvmin 2 which needs to be equal to the original potential energy mgh = 1 mv 2 + mg(2R) 2 min thus getting h = 2R + 2 vmin 5 = 2R + (1/2)R = R 2g 2 hence h = 45.25 m 9 3) So energy conservation dictates 1 5 mg R = mvb2 2 2 Solving for the speed √ 5gR vb = and vb = 29.796 m/s 4) 5 1 mg R = mvf2 + mgR 2 2 gives √ 3gR vf = so vf = 23.080 m/s 5) The kinetic energy will be transferred to the spring 1 1 mv 2 = kx2 2 f 2 Solving for distance √ mvf2 x= k √ = 3mgR k hence x = 1.594 m 6) So the initial energy is 1 mv 2 + mg(2R) 2 0 which must be equal to the energy at the top of the loop T Ei = TE = 1 mv 2 + mg(2R) 2 min Solving for the initial velocity v0 = vmin or v0 = 13.325 m/s 10