QUIZ 9 (ANSWERS) CH. 7: Problems 9*, 24*, 38* (TRY #34*, critical for Ch 12 and Physics 4B; see class notes on stable and unstable equilibrium) 41*, 42*, 43*, 46* (try #12*), 47* (TRY 65* and 66*) , 50*, 55*, 56*, 63* (see #46*) , 68*, 70*, 75* (SEE SAMPLE TEST 3) (ANSWERS) |
VIRTUAL LAB 2 (SEE EXAMPLE 7.7, CHAPTER 7) |
SIMPLE HARMONIC MOTION LAB |
Sample
Exam Cross Index for TEST 2 Sp '12
BELOW ARE THE LINKS TO SAMPLE EXAMS:
MEANWHILE HERE ARE LINKS (A), (B), (C) AND (D) TO
SAMPLE TEST 2 AND 3:
(A) http://www.nvaphysics.com/4A_ (B) http://www.nvaphysics.com/4A_ (C) http://www.nvaphysics.com/4A_ (D) http://www.nvaphysics.com/4A_ |
For many of these problems use KE_{i} + U_{i}
= KE_{f} + U_{f }+ HEAT or
USE KE_{i} + U_{i} + W_{F}
= KE_{f} + U_{f }+ HEAT, where U is the
potential energy due to the conservative force (
gravitational and spring forces in this case), KE is the
kinetic energy and W_{F} is the work done by
the external force (done by "Frankie" as we
discussed in class. ) NOTE: 9, 46 and 63 mix ENERGY CONSERVATION (Ch. 7)and CIRCULAR DYNAMICS (Ch. 5), a special class of problems you should study ahead of Test 2. ANOTHER NOTE: #50 mixes ENERGY CONSERVATION (Ch. 7) and PROJECTILE MOTION (CH. 3) |
9.* KE_{i} + U_{i} = KE_{f} + U_{f
}+ HEAT . (a) The angle between the normal force and motion in this case is 90 degrees, so what is the work by that force? The work by gravity is positive since the motion is from high to low; note the gravitational work only depends on the vertical displacement. (b) ENERGY CONSERVATION : KE_{i} + U_{i} = KE_{f} + U_{f }+ HEAT where HEAT IS THE NEGATIVE OF THE WORK BY FRICTION. THUS HEAT IS ALWAYS POSITIVE. You are given the heat. (c) CIRCULAR DYNAMICS: Critical point. HEAT is the negative of the work by the friction force, which may not be constant. The instantaneous friction force is u*N, where u is the kinetic friction coefficient and where N is the magnitude of the normal force which may not be constant as in this case. Clearly from earlier problems in Chapter 5, N is a maximum at the bottom and is a minimum at the top. (See worksheet 2, GRP 3, problem #1, THE PROBLEM DEALING WITH BUMPS AND VALLEYS) (d) CIRCULAR DYNAMICS: At bottom, the normal force points up in the pos direction toward the center of the circle and the weight of magnitude mg points down in the neg direction. Write: Sum of forces in radial direction = m*v^{2}/R = pos - neg and solve for N, the normal force magnitude, using the speed v you found in part (b) from energy conservation. See worksheet 2, GRP 3, problem #1. Note N > mg. |
24. *Use this: KE_{i} + U_{i}
= KE_{f} + U_{f }+ HEAT for the initial (i) and
final (f) locations. Study carefully the classic
example (i.e. Many physics book authors have used
it.). Unlike the previous problem, the friction force is
constant, so HEAT = f_{k}*D, where D is the distance
moved and f_{k }= 17,000 (N) is given. Note
the initial kinetic energy KE_{i} is not zero as in
problem 9. Set the lowest point equal to the zero of
gravitational potential energy (ie, y = 0 at point 2.) Note
U = U_{g} + U_{s}, where U_{g} = mgy
and U_{s} = (1/2)*k*y'^{2}, where y'
is the spring coordinate indicating compression. We have KE_{i}
= 16,000 J and U_{i} = mgy_{i} + 0
since the initial spring potential energy is zero (not
compressed--- y_{i}' = 0) ; note y_{i }=
2.00 m. U_{f }= mgy_{f} +
(1/2)*k*y_{f}'^{2 }, where y_{f} = 1.00 m
and y_{f}' = -1.00 m. (Since you square y' in general the negative sign goes away. ) The heat of course is f_{k}*D, where D = 1.00 m. Solve for KE_{f} and find final speed v_{f} : KE_{i} + U_{i} = KE_{f} + U_{f }+ HEAT means 16,000 J + mgy_{i} = KE_{f} + mgy_{f} + (1/2)*k*y_{f}'^{2 } + 17,000 (N)*d^{ }, where y_{f } = d = 1.00 m and y_{f}' = -1.00 m. (b) If you think the acceleration is constant think twice; the spring makes things more complex as we have already seen in Ch. 14. So use: sum of the forces in the y direction = m*a = pos - neg, where we set the positive y direction to be up. Thus: : m*a = | ky_{f}' | + f_{k} - mg, where clearly the spring and friction forces point upward as the object moves down. |
38.* READ section 7.4 and 7.5: If U = U(x) then F_{x} = -dU/dx. The x-component of force is the negative derivative of U; Thus if dU/dx is positive , the force points in the negative x -direction (F_{x }< 0) and if dU/dx is negative then force points in positive x direction (F_{x }> 0.) When dU/dx = 0 force is zero and the object is said to be in equilibrium, either stable or unstable. A CONCAVE UP POTENTIAL ENERGY FUNCTION MEANS EQUILIBRIUM IS STABLE. |
41*. (a) CHAPTER 5 REVIEW: You must prove whether or not
the system is at rest; the emphasis here is the use of Ch. 5
concepts, whereas Ch. 7 is more directly related to part
(b). If the system is at rest and remains so we have for the concrete m*a = pos - neg = mg - T = 0 . So you can immediately find T. For the Box we have m'*a = 0 = T - f_{s}
' , where f_{s} ' is the force of static friction from the
ground and m' is the Box mass. We also have m''*a =
0, where m'' is the gravel mass. ( i.e. There is no friction
force between the gravel and Box if the system is at rest.)
Compare T and f_{smax} = u_{s}*N, where N is
the magnitude of the normal force on the box, which must be equal
to (m' + m'')*g. |
42.* KE_{i} + U_{i} = KE_{f} + U_{f
}+ HEAT , where HEAT = 0.
(a) See
VIRTUAL LAB 2, question 1. |
43.* The simplest way to do the problem is to bypass the speed and
kinetic energy just after the Block leaves the spring
and consider only two points in space-time: When the spring is fully
compressed and the block is at rest (i) and when the block
has reached its maximum horizontal distance and comes to
rest permanently (f) . KE_{i} + U_{i} = KE_{f} + U_{f }+ HEAT , where U_{g} = 0 at both points (ground level, y = 0) . You should be able to see which terms are clearly zero. Thus we have initial (i) spring potential energy = HEAT , where HEAT = f_{k}*D : (1/2)*k*d^{2} = u*mg*D, where u = coefficient of kinetic friction and d and D are given in diagram. |
46.* ANOTHER CLASSIC, like #24. This one mixes energy
conservation and dynamics just like #9 above and #63 below. (a) KE_{i} + U_{i} = KE_{f} + U_{f }+ HEAT , where HEAT = 0. ENERGY CONSERVATION: (I) KE_{i} + U_{i} = KE_{f} + U_{f } , WHERE KE_{i} = 0, U_{i} = m*g*h, KE_{f} = (1/2)*m*v^{2} and U_{f} = m*g*2R: mgh = (1/2)*m*v^{2} + m*g*2R AT TOP. CIRCULAR DYNAMICS: ALSO AT TOP, the normal force and the weight of magnitude mg point down in the pos direction toward the center of the circle. Write: (II) Sum of forces in radial direction = m*v^{2}/R = pos - neg = N + mg - 0 = N + mg, since there is no force pointing in the neg direction away from the center; see #120 AND #42, Ch. 5 , point B at top. Set N = 0 IN THIS CASE. Solve equations I and II simultaneously for h in terms of R by eliminating v. The mass m cancels out. (b) FOR THIS INITIAL HEIGHT h, greater than that of part (a), the object never loses contact with the track at the top so it does reach point C. (I) ENERGY CONSERVATION : (I) KE_{i} + U_{i} = KE_{f} + U_{f } , WHERE KE_{i} = 0, U_{i} = m*g*h, h = 3.50*R, KE_{f} = (1/2)*m*v^{2} and U_{f} = m*g*R (at point C) . Find v at C: mg*( 3.50*R ) = (1/2)*m*v^{2}
+ m*g*R (III) For the tangential acceleration, let the pos direction
be vertically down, a direction tangent to the circle at point
C. |
47.* ANOTHER CLASSIC: m*g*H = TOTAL HEAT, SINCE THE INITIAL
AND FINAL kinetic energies are zero. The Heat is the total heat
after possibly numerous trips across the flat section. TOTAL HEAT
= f_{k}*L, where L is the total distance moved on flat surface before coming to rest. To find the number of trips, divide TOTAL HEAT by f_{k}*D, where D = 30 m exactly. From this ratio you can find the number of complete trips; from the fractional remainder, you can find exactly where the object comes to rest on the flat surface. If the ratio is less than one, then the object does not make a complete trip across the flat surface. |
50.* KE_{i} + U_{i} = KE_{f}
+ U_{f }+ HEAT , where HEAT = 0. See problem 68 hint
below. i = bottom. f = top just before leaving horizontal surface 70 (m)
above ground zero ( where U_{g} = 0) U_{f} = mgy_{f}, where y_{f} = 70 (m) KE_{i} = (1/2)*m*v_{i}^{2} . KE_{f} = (1/2)*m*v_{f}^{2}
. |
55.* KE_{i} + U_{i} = KE_{f}
+ U_{f }+ HEAT , where HEAT = 0.
Let the zero of gravitational energy be the ground. # Initially (i) the system is at REST and U_{i}
= the gravitational potential energy of the 12.0 kg clock at the
given initial height. Solve for the final common speed v_{f} of the two blocks. |
56.*
USE KE_{i} + U_{i} + W_{F}
= KE_{f} + U_{f }+ HEAT, where U is the
gravitational potential energy in this case, KE is the
kinetic energy, W_{F} is the work done by
the external thrust force and the HEAT is f_{k}*D. The
constant friction force of magnitude f_{k} = 500
(N) is given; D is the distance moved along ramp. # Initially (i), the rocket is at REST and U_{i} = the gravitational energy at the initial vertical height H above the ground. Note the height H has a simple relationship with D, the distance moved along the ramp; use simple geometry, the angle of 53 degrees, and the properties of right triangles to write H in terms of D. W_{F} = F*D, where D is the distance moved along the
ramp and F is the magnitude of the given external thrust force =
2000 (N) exactly. |
63.* ANOTHER CLASSIC: THE STRUCTURE OF THIS PROBLEM,
LIKE 9, and 46, mixes ENERGY CONSERVATION (Ch. 7)
and
CIRCULAR DYNAMICS (Ch. 5) ENERGY CONSERVATION (Ch. 7): (I) KE_{i} + U_{i} = KE_{f} + U_{f }+ HEAT , where HEAT = 0. # Initially (i), skier is at the top and is essentially at rest; KE_{i } = 0. U_{i }= m*g*R if the zero of potential energy is considered to be at the level of the snow ball's horizontal diameter. (I) mg*R = (1/2)*m*v_{f}^{2} + mg*R*cos alpha, where alpha is angle with vertical shown. # Finally (f), we consider the point when the skier leaves the snow ball , i.e. when the normal of magnitude N becomes zero. See figure 7.63. We define the location of this point in the following way: Draw a radial line from the snow ball center to this point and define the angle with the vertical as alpha . KE_{f} = (1/2)*m*v_{f}^{2} and U_{f} = m*g*R*cos alpha. CIRCULAR DYNAMICS: At any point on the surface of the ball, the normal force points radially away from the center of the circle. The gravitational force of magnitude mg points vertically down and the component of the gravitational force along the radial line is mg*cos theta and points toward the center of the circle. Write: (II) For the centripetal (center seeking) force, Sum of forces in radial direction = m*v^{2}/R = pos - neg = mg*cos theta - N. For this problem set N = 0. Thus: m*v_{f}^{2}/R = mg*cos theta. Solve equations I and II simultaneously for cos theta by eliminating v. Compute theta by evaluating cos ^{-1}. Your angle will be between 45 and 90 degrees. ^{ } |
68. WE DO THIS PROBLEM USING TWO STYLES--FROM CH. 6 AND
7. See #50 above. CH. 6 STYLE: (i) (1/2)*m*v_{f}^{2} - (1/2)*m*v_{i}^{2} = total work = W_{g} + W_{f} + W_{N}, work by gravity, the friction force and normal force respectively where last term is zero since the normal work on the way up slope must be vanish because that force is perpendicular to the motion i.e. makes an angle of 90 degrees. ALSO, THE SLOPE IS FRICTIONLESS MAKING WORK OF FRICTION W_{f} = 0. (ii) FIND THE SPEED AT THE TOP OF THE CLIFF USING (1/2)*m*v_{f}^{2} - (1/2)*m*v_{i}^{2} = total work = W_{g} WHERE THE NET VERTICAL DISPLACEMENT IS FROM LOW TO HIGH. IS GRAVITY WORK POSITIVE OR NEGATIVE? (ii) AFTER LEAVING CLIFF, DURING FREE FALL (i.e., PROJECTILE MOTION) THE gravitational work will be positive for this high to low motion. IN THAT PHASE OF PROBLEM YOU CAN USE CH. 3 METHODS AS IN LAB 3. HERE'S HOW YOU DO IT CH. 7 STYLE: |
70 (a) KE_{i} + U_{i} = KE_{f} + U_{f
}+ HEAT , where HEAT = 0. The block is released from rest
so that should tell you the value of KE_{i} . The maximum
speed occurs when U is a minimum and that location is shown in
figure 7.42. U_{f} = 0 assuming both springs are
un-deformed and the block is in equilibrium. U_{i} =
sum of two spring potential energies: Spring 1 is stretched by
0.15 m and spring 2 is compressed by 0.15 m. To evaluate U_{i},
apply the formula (1/2)*k*x^{2} for the
potential energy to both springs and add the two results.
Solve for KE_{f} and the maximum speed v_{f}
of the block WHICH OCCURS WHEN BOTH STRINGS ARE
UN-DEFORMED. (b) The simplest way to do this part is to assume the initial position (i) of the block is the same as in part (a) but the final location (f) is when Spring 1 is at maximum compression. When Spring 1 is at maximum compression, the block is at rest momentarily (i): U_{i} = U_{f }= sum of two spring potential energies. Spring 1 is compressed a distance |x'| and Spring 2 is stretched by the same |x'| . U_{i} was evaluated in part (a). Plug the symbol |x'| into the formula for U_{f} and solve for |x'| numerically. |
75. * SELECTIVELY USE KE_{i} + U_{i}
+ W_{F} = KE_{f} + U_{f }
where U is the potential energy due to the spring force in
this case, KE is the kinetic energy and W_{F}
is the work done by the external force of magnitude F. NO
FRICTION SO HEAT = 0. The FULL equation with W_{F} only applies between initial point A and point-B but we know the process continues beyond that. The Block will continue to move beyond point-B and eventually come to rest. At that point, the full distance from the origin is the amplitude A of subsequent simple harmonic motion about the origin and would be described by the function Acoswt for t >0 where t=0 is when the block reaches maximum distance from origin---see Ch. 14 and virtual lab 2.. So that is your task---to find that distance . Here are some tips: (a) Find the speed of the block at point-B using : KE_{A} + U_{A} + W_{F} = KE_{B} + U_{B}_{ } where W_{F} = F*D. Note KE_{A} = 0 and U_{A} = 0. Also W_{F} = F*D, D = 0.25 m and U_{B} = (1/2)*k*(0.25)^{2} in Joules. Find the final kinetic energy and final speed at B. (b) Find the additional distance before coming to rest using: KE_{B} + U_{B} = KE_{C} + U_{C}_{ } . NOTE: KE_{B } is known from the previous part and U_{B} = (1/2)*k*(0.25)^{2} Also KE_{C} = 0 and U_{C} = (1/2)*k*x_{C}^{2} . Find x_{C}. Then find 0.60 m - x_{C} . Note: KE_{A} + U_{A} + W_{F} = KE_{C} + U_{C } or, W_{F} = KE_{C} + U_{C}, where W_{F} = F*D, D = 0.25 m and KE_{C} = 0. |