Unformatted text preview: 178 CHAPTER 7 KINETIC ENERGY AND WORK CHAPTER 7 Answer to Checkpoint Questions 1. 2. 3. 4. 5. (a) decrease; (b) same; (c) negative, zero d, c, b, a (a) same; (b) smaller (a) positive; (b) negative; (c) zero zero Answer to Questions 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. all tie (a) same; (b) increasing (a) increasing; (b) same; (c) same; (d) increasing (a) yes (constant force); (b) and (c) no (variable force) (a) positive; (b) negative; (c) negative (a) 0; (b) 0 (a) positive; (b) zero; (c) negative; (d) negative; (e) zero; (f) positive (a) zero; (b) positive; (c) negative all tie (a) to (c) yes; (d) and (e) no c, d, a, and b tie, then f, e b (positive work), a (zero work), c (negative work), d (more negative work) (a) 3 m; (b) 3 m; (c) 0 and 6 m; (d) negative direction of x (a) 2F1 ; (b) 2W1 (a) A; (b) B (a) to (d) 3, 2, 1 CHAPTER 7 KINETIC ENERGY AND WORK 179 17. twice Solutions to Exercises & Problems (a) Let the mass of the system be m and the speed be v, then its kinetic energy is 1 K = 1 mv2 = 2 (2:9 105 kg)(11:2 103 m/s)2 = 1:8 1013 J : 2
2E 1E The kinetic energy is given by K = 1 mv2 , where m is the mass and v is the speed of the 2 electron. The speed is therefore v= r 2K = m s 2(6:7 10 19 J) = 1:2 106 m/s : 9:11 10 31 kg Use K = 1 mv2 . 2 (a) K = (110 kg)(8:1 m/s)2 =2 = 3:6 103 J: (b) K = (4:2 10 3 kg)(950 m/s)2 =2 = 1:9 103 J: (c) K = (91400 103 kg)[(32 knots)(0:515 m/s/knot)]2 =2 = 1:1 1010 J: (a) The change in kinetic energy for the meteorite is K = Kf Ki = Ki = 1 mi vi2 2 1 = 2 (4 106 kg)(15 103 m/s)2 = 5 1014 J ; where the negative sign indicates that kinetic energy is lost. (b) The energy loss in the unit of megaton TNT
14 5 K = 4:2 10 10 J6 TNT = 0:1 megaton TNT : 15 J=10
4E 3E 180 CHAPTER 7 KINETIC ENERGY AND WORK (c) The number of bombs the meteorite impact corresponds to is 10 TNT n = 0:1 103 TNT = 8 : 13
6
5E Let the diameter of the crater created by an explosion of energy E be d. Then d = CE 1=3 , where C is a constant. (a) The kinetic energy associated with the impact is d E= C 3 km 3 = 50km (1 magaton TNT) = 1 105 magaton TNT : 1
(b) The energy released from the Hiroshima bomb was equivalent to EH = 1:3 104 ton TNT. Let the energy associated with the impact be equivalent to that of n Hiroshima bombs. Then 5 magaton TNT n = 1 10 104 ton TNT = 1 107 : 1:3 (a) Let the initial speed of the proton be vi and the nal speed after going through a 2 distance d be vf . Then vf vi2 = 2ad; which yields
6P vf = vi2 + 2ad = (2:4 107 m/s)2 + (2)(3:6 1015 m/s2 )(0:035 m) = 2:9 107 m/s :
(b) The gain in kinetic energy is
2 k = 1 m(vf vi2 ) = 1 (1:67 10 27 kg)[(2:9 107 m/s)2 (2:4 107 m/s)2 ] 2 2 13 J)(1 eV=1:6 10 19 J) = 1:3 MeV : = (2:08 10 q p Let the mass of the man be m and his speed be vm . The kinetic energy of the man is 2 Km = Ks =2 = (m=4)vs =2, where vs is the speed of his son. Then 1 m(v + 1:0 m/s)2 = K = 1 m v2 = 2K = mv2 ; s22 s m m 2m
7P CHAPTER 7 KINETIC ENERGY AND WORK 181 1 v2 = (v + 1:0 m/s)2 m 2s >1 2 : v = v2 : 4s m The solution is vm = 2:4 m/s and vs = 4:8 m/s. Let the average EarthSun distance be D, then the average speed of the Earth is ve = 2D=T, where T = 1 y. If the mass of the Earth is denoted as Me , then the work that has to be done to stop the orbital motion of the Earth around the Sun would be
2 D2 2 W = 1 Me ve = 2 T 2 Me 2 2 11 m)2 (5:98 1024 = (2 )(1:5 10s/da)(365 da/y)]2 ) = 2:7 1033 J : [(1 y)(86400
8P i.e. 8 > < (a) The magnitude of the force required is 9E F = ma = (102 kg)(2:0 m/s2 ) = 2:0 102 N :
2 (b) The distance d traveled is obtain from v0 = 2ad: 2 2 v0 (53 d = 2a = 2(2:0m/s)2 ) = 7:0 102 m : m/s (c) The work done is W = Fd = (2:0 102 N)(7:0 102 ) = 1:4 105 J :
(d) For a = 4:0 m/s2 (double the original value), we have F = 2(2:0 102 N) = 4:0 102 N; 1 d = 2 (7:0 102 m) = 3:5 102 m ;
and W = Fd = (4:0 102 N)(3:5 102 N) = 1:4 105 J : 182 CHAPTER 7 KINETIC ENERGY AND WORK (a) The force exerted by the worker on the crate is constant so the work it does is given by WF = F d = Fd cos , where F is the force, d is the displacement of the crate, and is the angle between the force and the displacement. Here F = 210 N, d = 3:0 m, and = 20 . Thus WF = (210 N)(3:0 m)(cos 20 ) = 590 J. (b) The force of gravity is downward, perpendicular to the displacement of the crate. The angle between this force and the displacement is 90 and cos 90 = 0, so the work done by the force of gravity is 0. (c) The normal force of the oor on the crate is also perpendicular to the displacement, so the work done by this force is also 0. (d) These are the only forces acting on the crate so the total work done on it is 590 J. The work done by the water on the ice block is
11E 10E W = F d = [(210 N) i (1500 N) j] [(15 m) i (12 m) j] = 5:0 103 J :
12E The work W done by the force F is W = F d = [(2 N) i (4 N) j] [(8 m) i + c j] = 16 J (4c) N :
(a) Since W = 16 J (4c) N = 0, c = 4 m. (b) Since W = 16 J (4c) N > 0, c < 4 m. (c) Since W = 16 J (4c) N < 0, c > 4 m. The work done on the electron is equal to its gain in kinetic energy. Let the mass of the electron be m and its nal speed be vf , then the work done to accelerate the electron from zero speed to vf is 121 W = 2 mvf = 2 (9:1 10
31 kg)(3:0 106 m/s)2 =
13E 7:5 10 15 J = 47 keV : 1:6 10 19 J/eV From W = K we know that the work done by the force on the body is positive if its speed (and therefore kinetic energy) increases. Thus for AB, W is positive; for BC, W = 0; for CD, W is negative; and for DE, W is positive (note that jvj increases from C to D). 14E CHAPTER 7 KINETIC ENERGY AND WORK 183 The total mass of the rehose is m = (0:25 kg/m)(12 m) = 3:0 kg. Let the nal speed of the entire rehose be v, then the work done is 1 1 W = 2 mv2 = 2 (3:0 kg)(2:3 m/s)2 = 7:9 J : (a) The net force exerted on the trunk is Fnet = F1 + F2 + F3 , whose component along the displacement direction is Fh = F1 F2 cos 60 . Let the displacement of the trunk be d, then the work done by the net force is
16P 15E W = F d = Fh d = (F1 F2 cos 60 )d = [5:00 N 9:00 N(cos 60 )](3:00 m) = 1:5 J :
(b) Since W > 0 the kinetic energy of the trunk increases. The speed of the particle v as a function of time t is given by
17P d v(t) = dx = dt (3:0t 4:0t2 + 1:0t3 ) = 3:0 8:0t + 3:0t2 : dt
The inital and nal speeds of the particle are vi = 3:0 (8:0)(0) + (3:0)(0)2 = 3:0 m/s and vf = 3:0 (8:0)(4:0)+(3:0)(4:0)2 = 19 m/s, respectively. The work W done on the particle of mass m by the force F is equal to the gain in the kinetic energy of the particle. Thus 12 1 W = Kf Ki = 2 m(vf vi2 ) = 2 (3:0 kg)[(19 m/s)2 (3:0 m/s)2 ] = 5:3 102 J : (a) The x component of the net force F exerted on the canister is Fx = F3 cos 35 F1 F2 sin 50 , and the y component is Fy = F3 sin 35 F2 cos 50 . The magnitude of the net force is
2 F = Fx + Fy2 = (F3 cos 35 F1 F2 sin 50 )2 + (F3 sin 35 F2 cos 50 )2
p q p
18P = [10:0 N(cos 35 ) 3:00 N 4:00 N(sin 50 )]2 + [10:0 N(sin 35 ) 4:00 N(cos 50 )]2 = 3:8 N : Since it starts from rest, the canister will always move in the same direction as the constant resultant force F. Denote the diplacement of the canister as d, then the work done by the net force is W = F d = Fd = (3:8 N)(4:00 m) = 15 J: 184 CHAPTER 7 KINETIC ENERGY AND WORK The force F can be expressed in component form: work done is then 19P F = (F cos 150 ) i + (F sin 150 ) j: The W=
= Z rf Z
ri
F dr = Z xf 2:0 m xi Fx dx + Z yf yi Fy dy
4:0 m = (10 N)(cos 150 )(2:0 m) + (10 N)(sin 150 )( 4:0 m) = 37 J : 0 (10 N)(cos 150 ) dx + Z 0 (10 N)(sin 150 ) dy 20E (a) Let the weight of the roof be wr and its vertical displacement be h, then the work W done by the machine to lift the roof is W = wr h = (41; 000 ton)(2; 000 lb/ton)(4:0 in.)(1 ft=12 in.) = 2:7 107 ftlb :
(b) Let the weight of the car be wc and its vertical displacement be h0 , then the work W 0 done by Mrs. Maxwell to lift the car is W 0 = wc h0 = (3; 600 lb)(2:0 in.)(1 ft=12 in.) = 1:5 102 ftlb :
21E (a) let the force exerted by the worker on the crate be F and the displacement of the crate be d, then the work done by the worker is W1 = F d = Fd = (209 N)(1:5 m) = 314 J :
(b) Since the vertical displacement of the crate is h = d sin 25 = (1:5 m)(sin 25 ) = 0:63 m, the work done by the weight of the crate of mass m is W2 = mgh = (25:0 kg)(9:80 m/s2 )(0:63 m) = 155 J :
(c) Since it is perpendicular to the direction of motion of the crate, the normal force does not do any work on the crate. (d) The total work done on the crate is Wt = W1 + W2 = 314 J + ( 155 J) = 158 J : CHAPTER 7 KINETIC ENERGY AND WORK 185 (a) The component of the force of gravity exerted on the ice block (of mass m) along the incline is mg sin , where = sin 1 (0:91 m=1:5 m) is the angle of inclination for the inclined plane. Since the ice block slides down with uniform velocity, the worker must exert a force F up the inclined plane with a magnitude equal to mg sin , i.e., 22E F = mg sin = (45 kg)(9:80 m/s2 ) 0:91 m = 2:7 102 N : 1:5 m (b) The work done by the worker is W1 = (2:7 102 N)(1:5 m) = 4:0 102 J :
(c) The work done by the weight of the block is W2 = (45 kg)(9:80 m/s2 )(0:91 m) = 4:0 102 J :
(d) Since it is perpendicular to the direction of motion of the block, the normal force does not do any work on the block. (e) The work done by the net force Fnet must be zero, since Fnet itself is zero. You can check this by adding W1 and W2 above. (a) From the freebody diagram to the right we nd F =T and 2T mg = 0 : The magnitude of the force F needed to lift the canister is then 1 1 F = 2 mg = 2 (20 kg)(9:80 m/s2 ) = 98 N :
23E F T T mg (b) The free end of the string must be pulled down by a distance d0 = 2d = 2(2:0) cm = 4:0 cm. (c) Work done by you: W 0 = Fd0 = (98 N)(4:0 cm) = 3:9 J: (d) work done by mg: W = mgd = (20 kg)(9:80 m/s2 )(2:0 cm) = 3:9 J: 186 CHAPTER 7 KINETIC ENERGY AND WORK (From the above equations we see that W 0 + W=0. This is consistent with the fact that the speed, and therefore the kinetic energy, of the canister are constant.) (a) Let F be the magnitude of the force exerted by the cable on the astronaut. The force of the cable is upward and the force of gravity mg is downward. Furthermore the acceleration of the astronaut is g=10, upward. According to Newton's second law F mg = mg=10, so F = 11mg=10. Since the force F and the displacement d are in the same direction the work done by F is WF = Fd = 11mgd=10 = (11)(72 kg)(9:8 m/s2 )(15 m)=10 = 1:2 104 J. (b) The force of gravity has magnitude mg and is opposite in direction to the displacement. It does work Wg = mgd = (72 kg)(9:8 m/s2 )(15 m) = 1:1 104 J. (c) The total work done is WT = 1:2104 J 1:1104 J = 1:1103 J. Since the astronaut started from rest this must be her nal kinetic energy. (d) Since K = 1 mv2 her nal speed is 2
24P v= r 2K = m s (2)(1:0 103 J) = 5:3 m/s : 72 kg (a) Let F be the magnitude of the force of the cord on the block. This force is upward, while the force of gravity, with magnitude Mg, is downward. The acceleration is g=4, down. Take the downward direction to be positive. Then Newton's second law is Mg F = Mg=4, so F = 3Mg=4. The force is directed opposite to the displacement so the work it does is WF = Fd = 3Mgd=4. (b) The force of gravity is in the same direction as the displacement so it does work Wg = (c) The total work done on the block is WT = 3Mgd=4+Mgd = Mgd=4. Since the block starts from rest this is its kinetic energy after it is lowered a distance d. p p (d) Since K = 1 Mv2 , where v is the speed, v = 2K=M = gd=2 after the block is 2 lowered a distance d. Denote the displacement of the spelunker of mass m during each stage as d and the work done by the lifting force during each stage as Wi (i = 1; 2; 3). Apply the workenergy 2 theorem to each stage. For stage 1, W1 mgd = K1 = 1 mv1 , where v1 = 5:00 m/s. This 2 gives 12 1 W1 = mgd + 2 mv1 = (80:0 kg)(9:80 m/s2 )(10:0 m) + 2 (80:0 kg)(5:00 m/s)2 = 8:84 kJ :
26P 25P Mgd. CHAPTER 7 KINETIC ENERGY AND WORK 187 For stage 2, W2 mgd = K2 = 0, which gives W2 = mgd = (80:0 kg)(9:80 m/s2 )(10:0 m) = 7:84 kJ : 2 For stage 3, W3 mgd = K3 = 1 mv1 , which gives 2 12 1 W3 = mgd 2 mv1 = (80:0 kg)(9:80 m/s2 )(10:0 m) 2 (80:0 kg)(5:00 m/s)2 = 6:84 kJ :
27E R Use W = xxif F(x) dx. The work done is W=
= Z Z 8:0 m 2:0 m 0 0 F(x) dx F(x) dx +
Z 4:0 m = (10:0 N)(2:0 m) + (5:0 N)(2:0 m) + (0 N)(2:0 m) + ( 2:5 N)(2:0 m) = 25 J : From Fig. 734 in the text we see that the acceleration a varies as a linear function of x: a(x) = kx, where k = 2:5 s 2 : The work done is then
28E 2:0 m F(x) dx + Z 6:0 m 4:0 m F(x) dx + Z 8:0 m 6:0 m F(x) dx W= Z xf 1 = 2 (10 kg)(2:5 s 2 )(8:0 m)2 = 8:0 102 J :
29P xi F(x) dx = Z 8:0 m 0 ma dx = mk Z 8:0 m 0 1 x dx = 2 mkx2 f (a) To estimate W, the work done in question, you just have to estimate the area enclosed by the following curve and lines: y = F(x), y = 0, x = 1 m, and x = 3 m. A crude estimate gives W 7 J. You can probably do better than this. (b) Analytically, the work done is given by W= Z xf =3 m 3 m a = x dx = 9 Nm2 31 11 m m 1m = 6J: xi =1 m F(x) dx = Z 3m
1m a dx x2 188 CHAPTER 7 KINETIC ENERGY AND WORK (a) The graph shows F as a function of x if x0 is positive. The work is negative as the object moves from x = 0 to x = x0 and positive as it moves from x = x0 to x = 2x0 . Since the area of a triangle is 1 ( base)( altitude), the work done from x = 2 0 to x = x0 is 1 (x0 )(F0 ) and the work 2 done from x = x0 to x = 2x0 is 1 (2x0 2 x0 )(F0 ) = 1 (x0 )(F0 ). The total work is the 2 sum, which is 0. (b) The integral for the work is 30P F(x) F0 x0 F0 2x 0 x W= Z 2x0 0 F0 x 1 dx = F x2 x 0 2x x0 0 2x0 0 = 0: 31P The work done is W=
= = Z rf
ri Z rf
F dr
Z Z ri [(2x N)i + (3 N)j] [(dx)i + (dy)j] (2x N) dx +
yf yi xf 4m 3m 2 N) + (3y N) = (x 2m 3m xi (3 N) dy = 6J: (a) The force F(x) as a function of x is plotted in the next page. The work done is represented by the shaded area in the gure. Estimate the area to obtain W. You should get W 13 J : (b) Analytically, the work done is 32P W= Z xf = (10 N)(2:0 m)(1 e xi F(x) dx = Z 2:0 m 0 (10e x=2:0 m N) dx 2:0 m=2:0 m ) = 13 J : CHAPTER 7 KINETIC ENERGY AND WORK 189 F(x) (N)
10 5 0 x (m)
1 2 3 4 5 which gives (a) Let the velocity of the particle at x = 0 be v0 . The velocity v of the particle of mass R 2 m as a function of x satises the workenergy theorem: W = 0x Fx dx0 = 1 mv2 1 mv0 ; 2 2
0 33P K(x = 3:0 m) = Z x 1 = [(0)(2:0 m 1:0 m) + ( 4 N)(3:0 m 2:0 m)] + 2 (2:0 kg)(4:0 m/s)2 ) = 12 J : (b) The change in kinetic energy for the body is 2 K = 8 J 1 mv0 = 8 J 1 (2:0 kg)(4:0 m/s)2 = 8:0 J : 2 2 Let the location of the body be x when its kinetic energy is 8 J, then 0 12 F(x0 ) dx0 + 2 mv0 W= Fx dx0 0 0 2:0 m = 0 + ( 4 N)(x 2:0 m) = K = 8:0 J ;
0 0 0 Z x Fx dx0 = Z 2:0 m Fx dx0 + Z x which gives x = 4:0 m. (c) Since Fx > 0 for 0 < x < 1:0 m, the maximum kinetic energy of the body is attained at x = 1:0 m. Thus Z 1:0 m 12 Kmax = Fx dx + 2 mv0 = (2:0 N)(1:0 m) + 16 J = 18 J : 0 190 CHAPTER 7 KINETIC ENERGY AND WORK (a) The freebody diagram for the crate in its nal position is shown to the right. The horizontal component of the equation of motion for the crate is 34P y T
θ F mg x F T sin = 0 ;
while the vertical component is Solve for F: F cos mg = 0 :
4:00 m p = 797 N : F (12:0 m)2 (4:00 m)2 (b) Since there is no change in the crate's kinetic energy, the total work done on it is zero. (c) Let the length of the rope be l, then the work Wg done by the force of gravity is = mg tan = (230 kg)(9:80 m/s2 ) Wg = mgy = mg(l l cos ) = (230 kg)(9:80 m/s2 )[12:0 m p (12:0 m)2 (4:00 m)2 ] = 1:55 103 J : (d) Since it is always perpendicular to the direction of motion of the crate, the force T exerted by the rope does not do any work on the crate. (e) Since the total work as well as the work Wr done by the rope on the crate is zero, the work W you do on the crate should cancel with the work done by the force of gravity: W = Wg = 1:55 103 J: (f) Since F is a variable force, you cannot use W = F distance; which is valid only for constant forces. (a) As the cage moves from x = x1 to x = x2 the work done by the spring is given by 1 W = ( kx) dx = 2 kx2 = 1 k(x2 x2 ) ; 221 x1 x1 where k is the force constant for the spring. Substitute x1 = 0 and x2 = 7:6 10 3 m. The result is W = 1 (1500 N/m)(7:6 10 3 m)2 = 0:043 J. Notice that 15 N/cm was 2 converted to 1500 N/m. (b) Now substitute x1 = 7:6 10 3 m and x2 = 15:2 10 3 m. The result is now W = 1 (1500 N/m) (15:2 10 3 m)2 (7:6 10 3 m)2 = 0:13 J. Notice this is much more 2 work than in the rst interval. Although the displacements have the same magnitude the force is larger throughout the second interval.
Z
35E x2 x2 CHAPTER 7 KINETIC ENERGY AND WORK 191 Let the spring constant be k and the maximum stretch be x, then the work done is 1 1 W = 2 kx2 = 2 (100 N/m)(5:00 m)2 = 1:25 103 J : (a) Let the initial and nal velocities of the body of mass m be vi and vf , respectively. Then from W = Kf Ki we get
37P 36E W= Z xf 21 = 1 mvf 2 mvi2 2 1 2 = 2 (2:0 kg)[vf (8:0 m/s)2 ] : Solve for vf : vf = 6:6 m/s. (b) Let the position of the body in meters be xf , then xi F(x) dx = Z 4:0 3:0 ( 6x mN) dx = ( 3x2 J) 4:0 3:0 = 21 J W= Z xf Solve for xf : xf = 4:7. Here the suppressed unit of xf is meters. (a) When an additional weight of F = 240 N 110 N = 130 N is added to the spring, a further stretch of x = 60 mm 40 mm = 20 mm is resulted. Thus the spring constant is 130 N k = F = 20 mm = 6:5 N/mm : x The amount of stretch x1 for the rst case (with F1 = 110 N) is then 110 N x1 = F1 = 6:5 N/mm = 17 mm : k Thus the pointer will point at 40 mm 17 mm = 23 mm upon removal of the weight. (b) The amount of stretch x3 with a weight w attached is x3 = 30 mm 23 mm = 7 mm. The weight w is then w = kx3 = (6:5 N/mm)(7 mm) = 45 N:
38P 21 = 1 mvf 2 mvi2 2 1 = 2 (2:0 kg)[(5:0 m/s)2 (8:0 m/s)2 ] : xi F(x) dx = Z xf 3:0 ( 6x mN) dx = ( 3x2 J) = ( 3:0 xf 3x2 + 27) J f 192 CHAPTER 7 KINETIC ENERGY AND WORK The situations of the system before and after the short cord is cut are depicted in the two gures below.
before after 50N 80cm 10cm 60cm 50N 50N 39P 100N 70cm 100N 70cm 50N 100N 100N (a) and (b) Before the cord is cut each spring is subject to a force F = 100 N, so each stretches by x = F=k = 100 N=(500 N/m) = 20 cm, bringing the total length of each spring to 50 cm+20 cm = 70 cm. The box is therefore located a distance (70 cm+10 cm+70 cm) = 150 cm below the ceiling. After the cord is cut, each spring (and cord) supports half of the weight of the box, so the new amount by which each spring now stretches is x0 = x=2 = 10 cm, bringing the new total length of each spring to 50 cm + 10 cm = 60 cm. Add that to the length of each cord (= 85 cm) and we nd that the box is now located a distance (60 cm + 85 cm) = 145 cm below the ceiling. Taking the dierence in the position of the box between the two cases, we nd that the box will move up by d = (150 cm 145 cm) = 5:0 cm after the cord is cut. (c) During the upward motion of the box the negative work done by the force of gravity on the box of mass m is Wg = mgd = (100 N)(0:050 m) = 5:0 J. Since there is no change in the kinetic energy of the box, this negative work must be oset by the positive work Ws done by the two springs. Therefore Ws = Wg = 5:0 J. (a) Let the compression of the spring be x. The work done by the block's weight is
40P W1 = mgx = (250 10 3 kg)(9:80 m/s2 )(12 10 2 m) = 2:9 10 1 J :
(b) The work done by the spring is W2 = 1 kx2 = 1 (2:5 102 N/m)(12 10 2 m)2 = 1:8 J : 2 2 CHAPTER 7 KINETIC ENERGY AND WORK 193 (c) The speed vi of the block just before it hits the spring is given by K = 0 1 mvi2 = W1 + W2 ; 2 which yields vi = r ( 2)(W1 + W2 ) = s m ( 2)(0:29 J 1:8 J) = 3:5 m/s : 250 10 3 kg (d) Le...
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 Fall '07
 GROUPTEST
 Energy, Kinetic Energy, Work

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