HW 3. Due: Tuesday, December 4, 2018, 6:00 p.m.

Transcription

1 Oregon State University PH 11 Fall Term 18 HW 3 Due Tuesday, December 4, 18, 6 p.m. Print your full LAST name Print your full first name Print your full OSU student ID# Turn this assignment in to box #15 (located outside of Wngr 34) OR a pdf document of the complete file, prepared as directed on the next page. I affirm and attest that this HW assignment is my own work. All the reasoning and results here are my own doing. Sign your name (full signature) Print today s date

2 Oregon State University PH 11 Fall Term 18 HW 3 Directions Solve these four problems fully, just as you would if they were exam problems. Then turn in your entire solutions set (all these pages) to Box 15; OR a.pdf file to coffinc@physics.oregonstate.edu with a subject line as follows HW3_LastName_FirstName Parts of your four solutions will be chosen and scored for HW3 credit (35 points). Full solutions to all four problems will be posted soon after the due date/time. Note You ll do much better with these problems if you work through the Prep 9-1 sets and also the SMT3 first, because these problems are all variations on certain Prep and SMT3 problems. Note also The real MT3 (that s the final exam) may also include variations on Prep and SMT3 problems.

3 1. Refer to the diagram. A platform (m P ) is designed to test various equipment for durability for a future mission to Mars. The platform can be accelerated by a powerful piston so that it slides along a ramp inclined at angle q. To simulate space conditions, the whole apparatus (the ramp, platform and piston) is housed in a long airless tube (not shown). But all the testing is done here on the earth s surface, where g = 9.8 m/s. piston platform q ramp Before testing any equipment, of course, the researchers must test the platform-piston assembly itself, as follows The platform is initially at rest. Then at t = s, it is pushed up the ramp by the piston (whose force is parallel to the ramp). The platform s acceleration (along the ramp surface) is a(t) = k( t) + 8, where a is in m/s and t is in s. The other data for this test m P = 3.6 kg q = 17. µ S.RP =.476 µ K.RP =.53 k = 4.15 m s 5/ a. For this test Calculate the (instantaneous) power being delivered to the platform by the piston at t =.1 s. b. For this test Calculate the total average power that the platform has received from all external forces (so exclude the gravitational force) during the entire interval t.1 s. a. P mech.piston.p (t) = F Piston.P (t)v P (t) Define the ramp surface as the x-axis (positive uphill). FBD of platform (m P ) SF P.x = m P a P.x SF P.y = m P a P.y y Now find v P (t) x F K.RP F N.RP F G.EP Evaluate at t =.1 s F Piston.P F Piston.P F K.RP F G.EP.x = m P a P.x F N.RP F G.EP.y = m P a P.y F Piston.P m K.RP (F N.RP ) m P gsinq = m P [k( t) + 8] F N.RP m P gcosq = Substitute F Piston.P m K.RP (m P gcosq) m P gsinq = m P [k( t) + 8] Thus F Piston.P (t) = m P [k( t) + 8] + m K.RP (m P gcosq) + m P gsinq So F Piston.P (.1 s) = m P [k(.1) + 8] + m K.RP (m P gcosq) + m P gsinq = (3.6)[(4.15)(.1) + 8] + (.53)[(3.6)(9.8)cos17 ] + (3.6)(9.8)sin17 = N v P (t) = v P () + a P (t) dt = + [k( t) + 8]dt = (k/3)t 3/ + 8t v P (.1 s) = [(4.15)/3](.1) 3/ + 8(.1) = 5.19 m/s Find P mech.piston.p (.1 s) P mech.piston.p (.1 s) = [F Piston.P (.1 s)][v P (.1 s)] t = (69.686)(5.19) = 176 W t pts. 1/ pt. b. By definition P mech.ext.p.avg = W ext.p /Dt And we know E mech.f That is But U S.i = U S.f = And U G.i = = W ext Therefore W ext = (1/)(m P )v P.f + m P gh f 1/ pt. where h f = x P.f (sinq) t t Now find x P (t) x P (t) = x P () + v P (t) dt = + [(k/3)t3/ + 8t]dt = (4k/15)t 5/ + 4t Evaluate x P.f x P (.1 s) = [4(4.15)/15](.1) 5/ + 4(.1) = 4.71 m Therefore P mech.ext.p.avg = [(1/)(m P )v P.f + m P gx P.f (sinq)]/dt = [(1/)(3.6)(5.19) + (3.6)(9.8)(4.71)(sin17 )]/.1 = 67 W pts. 1/ pt. 1/ pt.

4 . You step into an elevator on the 44th floor of a tall office building. Each floor is 3.5 m above the one below it. You re alone in the elevator. You select your destination floor, then as you wait for the door to close you notice There s a bathroom scale with a mass of.43 kg resting on the elevator floor. (Of course.) So you stand on the scales during the entire ride. (Of course.) Shown here is a time graph of the scale s reading of your weight during the ride. Weight (N) t (s) The elevator is guided along vertical rails but is supported essentially by a single cable. F T (N) 14,56 Shown here is a time graph of that cable s tension, F T, during the ride. 11,996 11,9 1, (For simplicity, treat the ends of both of these graphs as infinite-slope discontinuities.) Assume g = 9.8 m/s. Use an energy analysis to find the average total resisting force (which is probably a combination of friction by the rails and air drag) that was exerted on the elevator during your ride. (You may be able to confirm or check your result by other methods, too good idea but use energy for your main analysis.) t (s) For the whole ride That is U S.i = U S.f = = More detail E mech.f U G.f = W ext mgh f = W T + W R mgh f = P T (t) dt + F R.avg Dy mgy f = [F T1 (t) v 1 (t)]dt + [F T (t) v (t)]dt + [F T3 (t) v 3 (t)]dt + F R.avg y f Now note here All forces are upward (+y direction); all velocities and displacements are downward ( y). Therefore F T (t) v(t) = F T (t)v(t), and F R.avg y f = F R.avg y f. And therefore Solving for F R.avg pts. mgy f = [F (t)v (t)]dt + [F (t)v (t)]dt + [F (t)v (t)]dt + F y T1 1 T T3 3 R.avg f F R.avg = mg (1/y f ){ [F (t)v (t)]dt + [F (t)v (t)]dt + [F (t)v (t)]dt} T1 1 T T3 3 pts. The elevator s motion from the solution to HW, problem 3... Interval 1 ( s t 5 s) a 1 = (1.4/5)t 1.4 =.8t 1.4 v 1 (t) =.14t 1.4t Interval (5 s t 16 s) a = v (t) = 3.5 m/s v (16) = v (5) = 3.5 m/s Interval 3 (16 s t s) a 3 = (1.75/4)(t 16) =.4375t 7 v 3 (t) =.1875t 7t And y 3 () = y f = 59.5 m

5 From the graph of F T F T1 (t) = (17/5)t + 1, = 34t + 1, N F T (t) = 11,9 N F T3 (t) = (156/4)t + 1, = 539t + 3,76 N mg = 11,996 N 1/ pt. 1/ pt. Substituting 5 F R.avg = 11,996 + (1/59.5){ [(34t + 1,)(.14t 1.4t)]dt 16 + [(11,9)( 3.5)]dt 5 + [(539t + 3,76)(.1875t 7t + 5.5)]dt } 16 = 94. N

6 3. Refer to this diagram A B Mass A (6.3 kg) is attached to the left end of an ideal (massless) spring that has a stiffness of 7.4 N/m. Mass B (4.56 kg) is not attached to anything. Both masses are supported by (and can slide freely on) a level, frictionless surface. There is no air drag nor any other resistance losses. Initially Mass A is at rest; the spring is relaxed; and mass B is moving to the left (toward mass A) at 1.53 m/s. Then mass B collides with the right (free) end of the spring. Sometime later, mass B ceases contact with the spring. Assume that no mechanical energy is lost during the collision (i.e. that this is a perfectly elastic collision) because the spring is essentially massless. a. How far is the spring is compressed when the two masses (and spring) are moving at a single common speed? b. Find the velocity of mass A after mass B is no longer in contact with the spring. c. Now refer instead to this diagram, where there are two changes to the above scenario. A B Now the surface under mass A is not frictionless (but the surface under B is still frictionless) Now the spring is still ideal (perfectly elastic) but not a linear restoring force (does not exhibit Hooke s Law). Rather, the force it exerts when compressed by a distance x is given by F = cx 3/, where c = 18.7 N m 3/ With those two changes, suppose that the collision experiment is repeated (same masses; same initial velocity of mass B). Assuming g = 9.8 m/s, what is the minimum coefficient of static friction that must exist between mass A and the surface under it so that A does not move during B s collision with the spring? a. First, use the conservation of (x-) momentum principle to analyze this collision from the moment shown above (top) to the moment when the two masses have equal speeds. (Let +x be defined to the right here.) The principle But = v B.f = v f And v A.i = Solving for v f SP x.i = SP x.f m A v A.i v B.f m A v A.i = (m A )v f = (m A )v f 1/ pt. v f = /(m A ) = (4.56)( 1.53)/( ) =.6466 m/s Next, use the additional given constraint that the total mechanical energy of the system remained the same (i.e. that no external work was done on the system) throughout the above interval The principle U S.i = U G = W ext = E mech.f Therefore (1/)(m A )v f + (1/)kx f = (1/) Solving x f = {[ (m A )v f ]/k} 1/ 1/ pt. 1/ pt. 1/ pt. 1/ pt. 1/ pt. = {[(4.56)(1.53) ( )(.6466) ]/7.4} 1/ =.474 m 1/ pt. 1/ pt.

7 b. Now use the same two principles as above (same momentum and same mechanical energy but this time over the entire collision (from mass B s first contact to last contact with the spring). Again, define +x to the right. Momentum SP x.i = SP x.f m A v A.i v B.f I. v A.i = v B.f Mechanical energy U S = U G = W ext = E mech.f II. Therefore (1/)m A + (1/) v B.f = (1/) Solve the above two equations (I and II) simultaneously for Solve I for v B.f v B.f = ( m A )/ Substitute into II (1/)m A + (1/) [( m A )/ ] = (1/) Expand (1/)m A + (1/) [( m A + m A )/ ] = (1/) Simplify m A m A + m A = Collect (m A + m A ) (m A ) = So, either =, or = (m A )/(m A + m A ) = [(6.3)(4.56)( 1.53)]/[(6.3)(4.56)+(6.3) ] = 1.9 m/s (We known, because block A was pushed to the left by an unopposed force from the spring; it must have accelerated in that direction from its initial velocity, which was zero.) c. If the friction holds block A in place (barely), then there is a moment when it halts B s motion before reversing it so at that moment, both blocks have zero speed; all energy is in the compression of the spring Mechanical energy U S.i = U G = W ext = E mech.f U S.f But what is U S.f here? We re dealing with a non-linear ideal spring. Derive U S U S = [F SM (x)]dx = [ cx3/ ]dx = cx3/ dx x = (/5)cx 5/ x x Therefore (/5)cx f 5/ = (1/) Solving x f = [(5/4)( )/c] /5 = [(5/4)(4.56)( 1.53) /18.7] /5 = m 1/ pt. Block A at F S.SA max SF A.x a A.x SF A.y a A.y F N.SA y max F S.SA F Spr.A a A.x F N.SA F G.EA a A.y min 3/ m S (F N.SA ) cx f = 1/ pt. F N.SA m A g = 1/ pt. x F Spr.A F S.SA max min 3/ Thus m S (m A g) cx f = 1/ pt. min min 3/ Solving for m S m S = cx f /(m A g) = (18.7)(.87371) 3/ /[(6.3)(9.8)] =.5 1/ pt. 1/ pt. F G.EA

8 4. (More practice with energy graphs...) a. The graph here shows the total potential energy that a certain object (m = 4. kg) would have along the r-axis. Assuming that no external work is done (i) What minimum speed must the object have at r = 4. m in order to reach r = 14. m? U (J) (ii) Assuming the result of step (i) above,. find the object s speed at r = 14. m. r (m) (iii) Now assume that the actual speed of the object at r = 4. m is 1% greater than the value you calculated in step (i) above. Use two different methods to find the average r-acceleration of the object as it travels from r = 1. m to r = 1. m. (i) In order to travel to r = 14. m (starting at r = 4. m), the object would need to have enough total energy (E mech.total ) first to get past r = 1. m, which would require at least E mech.total = 1 J. Assuming that no external work is done (problem does not mention any), that s the same E mech.total the block will have at r = 4. m. At r = 4., the object has only 4 J of potential energy, so it will need at least 6 J of kinetic energy to have the required minimum E mech.total = 1 J. So, at r = 4. m (1/)mv min = 6 Solve for v min v min = [( 6)/m] = 3 = 5.48 m/s 1/ pt. 1/ pt. pts. pts. (ii) We already know E mech.total = 1 J. And at r = 14., the object has 8 J of potential energy, so it has J of kinetic energy. So, at r = 14. m (1/)mv = Solve for v v = [( )/m] = 1 = 3.16 m/s (iii) Now we assume v = 1.1v min at r = 4. m. That is And therefore v = = 6.5 m/s So, at r = 1. m (1/)mv = = 3.6 Solve for v E mech.total = 4 + (1/)(4)(6.5) = 11.6 J v = [( 3.6)/m] = 16.3 = m/s And at r = 1. m (1/)mv = = 1.6 Solve for v v = [( 1.6)/m] = 6.3 =.51 m/s Now, simply observing that the slope in the region 1. m r 1. m is constant, and knowing also that F r = du/dr, this tells us that F r (and therefore a r = F r /m) is also constant. So we can use constant-a kinematics here. One such route Solve for Dt Dr = (1/)(v i + v f )Dt Dt = Dr/(v i + v f ) = (.)/( ) =.619 s Then by definition a avg = (v f v i )/Dt = ( )/.619 =.5 m/s Alternatively, we can do this F r.avg = (DU/Dr) For 1. m r 1. m F r.avg = (1 8)/(1 1) And then a r.avg = F r.avg /m = 1 N (for 1. m r 1. m) = 1/4 =.5 m/s (for 1. m r 1. m)

9 b. Refer to the diagram (not to scale) Starting from rest, a block of mass m slides down a slope angled at q below the horizontal. There is kinetic friction (µ K ) between the block and the slope. There is also a steady wind blowing horizontally to the left, producing a steady (unknown) air drag force F D, as indicated. When the block has traveled a distance d to the bottom of the slope, it has a speed v. Known data m = 7.8 kg m K =.1 q = 19.5 d = 4.36 m v =.34 m/s Assumptions Define the x-axis as parallel to the sloped surface, with the origin at the block s starting point. Let U G = at the bottom of the slope. Assume g = 9.8 m/s. Show all reasoning and calculations for each step below. (i) Use axes such as those below to draw and label the graphs of E mech.total and U G (both on the same axes). For two points on each graph (where x = and x = d), calculate and label the values of each graph. Analyze the whole trip for E mech (initial point is at x = ; final point is at x = d = 4.36) The principle But U S = U G.f = Thus E mech.f = U G.i E mech.i = U G.i = mgh i = mgdsinq And E mech.f = = (1/)mv f Moreover And So below is the graph U G = = (7.8)(9.8)(4.36)sin19.5 = = 111 J m K = (1/)(7.8)(.34) = 1.35 = 1.4 J The graph of U G (x) must be a linear function of x; for any point x along the motion, U G (x) = mg h(x) = mg [(4.36 x)sinq]. The graph of E mech (x) must be linear, too, because W ext (x) is linear W ext (x) = [F K + F D.x ] x (and F K and F D.x are constants all during the motion). m F D q d E mech (J) U G (x) E mech.total (x) W ext (x) K T (x) 4.36 x (m)

10 (ii) Calculate the slope of your graph of U G. Explain how this result makes sense. Directly from the graph du G /dx = (U G.f U G.i )/Dx = ( 111.5)/4.36 = J/m = 5.5 N But we also know du G /dx = F G.x = mgsinq = (7.8)(9.8)sin19.5 = 5.5 N ] (iii) Calculate the slope of your graph of E mech.total. Explain how this result makes sense. Directly from the graph de mech.total /dx = (E mech.f E mech.i )/Dx But we also know E mech (x) (x) = ( )/4.36 =.619 J/m =.6 N = 111 [F K + F D.x ] x And so de mech /dx = [F K + F D.x ] This makes sense because F K and F D.x are the two external forces that do work on the block (and as observed previously, they are constant throughout the motion). (iv) Explain how your graphs show that the block accelerated as it moved down the sloped surface. K T (x) is represented on the graph by the difference (the gap ) between E mech (x) and U G (x), as noted above. That gap increases with x; the block gains speed during its motion.