(b) The mechanical energy would be 20% of the results of part (a), so (0 20)(920 m) 180 m.

Transcription

1 PH Chapter 7 Solutions 7.4. IDENTIFY: The energy from the food goes into the increased gravitational potential energy of the hiker. We must convert food calories to joules. SET P: The change in gravitational potential energy is grav mg ( yf yi ), while the increase in kinetic energy is negligible. Set the food energy, epressed in joules, equal to the mechanical energy developed. EXECTE: (a) The food energy equals mg( yf yi), so (40 food calories)(486 J/ food calorie) yf yi 90 m. (65 kg)(980 m/s ) (b) The mechanical energy would be 0% of the results of part (a), so y (00)(90 m) 80 m. EVALATE: Since only 0% of the food calories go into mechanical energy, the hiker needs much less of climb to turn off the calories in the bar IDENTIFY and SET P: se energy methods. Points and are shown in Figure 7.5. (a) W other. Solve for and then use mv to obtain v. Figure 7.5 Wother 0 (The only force on the ball while it is in the air is gravity.) mv ; mv mgy, y.0 m mgy 0, since y 0 for our choice of coordinates. EXECTE: mv mgy mv v v gy ( 0 m/s) (980 m/s )(0 m) 4 0 m/s EVALATE: The projection angle of 53 doesn t enter into the calculation. The kinetic energy depends only on the magnitude of the velocity; it is independent of the direction of the velocity. (b) Nothing changes in the calculation. The epression derived in part (a) for v is independent of the angle, so v 4 0 m/s, the same as in part (a). (c) The ball travels a shorter distance in part (b), so in that case air resistance will have less effect IDENTIFY: We treat the tendon like a spring and apply Hooke s law to it. nowing the force stretching the tendon and how much it stretched, we can find its force constant.

2 SET P: se Fon tendon k. In part (a), F on tendon equals mg, the weight of the object suspended from it. In part(b), also apply k to calculate the stored energy. EXECTE: (a) F (b) on tendon 38 N el Fon tendon (050 kg)(980 m/s ) k 99 N/m. 003 m 0.693m 69.3 cm; k 99 N/m el (99 N/m)(0.693 m) 47.8 J. EVALATE: The 50 g object has a weight of.45 N. The 38 N force is much larger than this and stretches the tendon a much greater distance IDENTIFY: The spring force is conservative but the force of friction is nonconservative. Energy is conserved during the process. Initially all the energy is stored in the spring, but part of this goes to kinetic energy, part remains as elastic potential energy, and the rest does work against friction. SET P: Energy conservation: W other, the elastic energy in the spring is k, and the work done by friction is Wf fks k mgs. EXECTE: The initial and final elastic potential energies are and k mv k (840 N/m)( m) J (840 N/m)(0 000 m) J. The initial and final kinetic energies are 0 and. The work done by friction is W W f s mgs (040)( 50 kg)(98 m/s )(0000 m) 0 96 J. Energy conservation gives other fk k k other mv W 0378 J ( 096 J) 0040 J 0 40 J. Solving for v gives (040 J) v 0335 m/s. m 50 kg EVALATE: Mechanical energy is not conserved due to friction IDENTIFY: Some of the initial gravitational potential energy is converted to kinetic energy, but some of it is lost due to work by the nonconservative friction force. SET P: The energy of the bo at the edge of the roof is given by: E mech, f E mech, i f k s. Setting y f 0 at this point, y i (45 m) sin36 50 m Furthermore, by substituting i 0 and conservation equation, f i k mv mgy f s or vf gyi fksg/ w g( yi fks/ w). EXECTE: vf (980 m/s ) (50 m) (0 N)(4 5 m)/(85 0 N) 54 m/s. f f mv into the EVALATE: Friction does negative work and removes mechanical energy from the system. In the absence of friction the final speed of the toolbo would be 700 m/s IDENTIFY: Apply Eq. (7.6). SET P: The sign of F indicates its direction. d EXECTE: F 4 (48 J/ m ). F ( m) (4.8 J/m )( 0.80 m).46 N. The force d is in the -direction. F when 0 and F 0 when 0, so the force is always directed towards the origin. EVALATE: 0

3 7.39. IDENTIFY and SET P: se Eq. (7.7) to calculate the force from. At equilibrium F 0. (a) EXECTE: The graphs are sketched in Figure a b r r 6 d a 6b F dr 3 7 r r Figure 7.39 (b) At equilibrium 0, d F so 0 dr F 0 implies a 6b r r 6 6 6br a; solution is the equilibrium distance r0 ( a/b) / is a minimum at this r; the equilibrium is stable. /6 6 (c) At r ( a/ b ), a/ r b/ r a( b/ a) b( b/ a) b /4 a. At r, 0 The energy that must be added is /6 0 (d) r0 ( a/ b ) 30 m gives that 60 6 ab / 08 0 m and b/4a 40 0 m 8 b /4 a b( b/4 a ) 54 0 J b /4 a b(40 0 m ) 54 0 J and b 640 J m Then ab / 08 0 m gives a ( b /)(08 0 m ) (640 J m ) (08 0 m ) J m EVALATE: As the graphs in part (a) show, Fr () is the slope of r () at each r. r () has a minimum where F IDENTIFY: Apply F ma to the bag and to the bo. Apply Eq. (7.7) to the motion of the system of the bo and bucket after the bag is removed. SET P: Let y 0 at the final height of the bucket, so y 00 m and y The bo and the bucket move with the same speed v, so ( m bo m bucket ) v. W other f k d, with d 00 m and f m g Before the bag is removed, the maimum possible friction force the roof can eert on the bo is k k bo. (0700)(80 0 kg 500 kg)(980 m/s ) 89 N. This is larger than the weight of the bucket (637 N), so before the bag is removed the system is at rest.

4 EXECTE: (a) The friction force on the bag of gravel is zero, since there is no other horizontal force on the bag for friction to oppose. The static friction force on the bo equals the weight of the bucket, 637 N. (b) Eq. (7.7) gives mbucket gy fkd m totv, with mtot 45 0 kg. v ( m bucket gy k m bo gd). m v [(650 kg)(980 m/s )(00 m) (0400)(80 0 kg)(980 m/s )(00 m)]. 450 kg v 99 m/s. EVALATE: If we apply F ma to the bo and to the bucket we can calculate their common acceleration a. Then a constant acceleration equation applied to either object gives v 99 m/s, in agreement with our result obtained using energy methods IDENTIFY: se the work-energy theorem, Eq. (7.7). The target variable k will be a factor in the work done by friction. SET P: Let point be where the block is released and let point be where the block stops, as shown in Figure W other tot Work is done on the block by the spring and by friction, W W and el. so other f Figure 7.43 EXECTE: 0, el, el 0, k (00 N/m)(000 m) 00 J since after the block leaves the spring has given up all its stored energy other f k k k W W ( f cos ) s mg cos s mgs, since 80 (The friction force is directed opposite to the displacement and does negative work.) Putting all this into Wother gives, el Wf 0 kmgs, el, el.00 J k 04. mgs (050 kg)(980 m/s )( 00 m) EVALATE:, el Wf 0 says that the potential energy originally stored in the spring is taken out of the system by the negative work done by friction IDENTIFY: The mechanical energy of the roller coaster is conserved since there is no friction with the track. We must also apply Newton s second law for the circular motion. SET P: For part (a), apply conservation of energy to the motion from point A to point B: with 0. Defining y 0 and y 30 m, conservation of energy B grav,b A grav,a becomes mv B mgya A B or v gy. In part (b), the free-body diagram for the roller coaster car at point B B A A

5 is shown in Figure Fy may gives mg n marad, where a v r Solving for the normal force v gives n m g. r rad /. Figure 7.45 EXECTE: (a) vb (980 m/s )(3 0 m) 6 0 m/s. (60 m/s) 4 (b) n (350 kg) 980 m/s 5 0 N. 60 m EVALATE: The normal force n is the force that the tracks eert on the roller coaster car. The car eerts a force of equal magnitude and opposite direction on the tracks (a) IDENTIFY: se work-energy relation to find the kinetic energy of the wood as it enters the rough bottom. SET P: Let point be where the piece of wood is released and point be just before it enters the rough bottom. Let y 0 be at point. EXECTE: gives mgy 78 4 J. IDENTIFY: Now apply work-energy relation to the motion along the rough bottom. SET P: Let point be where it enters the rough bottom and point be where it stops. Wother W W mgs 0; 78 4 J EXECTE: other f k, 784 J mgs 0; solving for s gives s 0 0 m. k The wood stops after traveling 0.0 m along the rough bottom. (b) Friction does 78 4 J of work. EVALATE: The piece of wood stops before it makes one trip across the rough bottom. The final mechanical energy is zero. The negative friction work takes away all the mechanical energy initially in the system IDENTIFY: se the work-energy theorem, Eq. (7.7). Solve for and then for v. SET P: Let point be at his initial position against the compressed spring and let point be at the end of the barrel, as shown in Figure se F k to find the amount the spring is initially compressed by the 4400 N force. W other Take y 0 at his initial position. EXECTE: 0, W W fs other fric mv Wother (40 N)(40 m) 60 J

6 Figure 7.53, grav 0, F kd so, el, el kd, where d is the distance the spring is initially compressed. F 4400 N d 4 00 m k 00 N/m and (00 N/m)(400 m) 8800 J, grav mgy (60 kg)(980 m/s )(5 m) 470 J,, el 0 Then Wother gives 8800 J 60 J mv 470 J (770 J) mv 770 J and v 5 5 m/s 60 kg EVALATE: Some of the potential energy stored in the compressed spring is taken away by the work done by friction. The rest goes partly into gravitational potential energy and partly into kinetic energy IDENTIFY: Initially the ball has all kinetic energy, but at its highest point it has kinetic energy and potential energy. Since it is thrown upward at an angle, its kinetic energy is not zero at its highest point. SET P: Apply conservation of energy: f f i i. Let yi 0, so yf h, the maimum height. At this maimum height, v f, 0 and vf, v i,, so vf v i, (5 m/s)(cos60.0 ) 7.5 m/s. Substituting into y conservation of energy equation gives mv i mgh m (7 5 m/s). vi (75 m/s) (5 m/s) (75 m/s) EXECTE: Solve for h: h 86 m g (980 m/s ) EVALATE: If the ball were thrown straight up, its maimum height would be.5 m, since all of its kinetic energy would be converted to potential energy. But in this case it reaches a lower height because it still retains some kinetic energy at its highest point IDENTIFY: Apply Eq. (7.4) to the initial and final positions of the truck. SET P: Let y 0 at the lowest point of the path of the truck. W other is the work done by friction. fr rn rmg cos. EXECTE: Denote the distance the truck moves up the ramp by. mv 0, mglsin, 0, mg sin and Wother - rmg cos. From Wother ( ) ( ), and solving for, 0 mglsin ( v / g) Lsin mg sin cos sin cos EVALATE: increases when v 0 increases and decreases when r increases. r r 7.8. IDENTIFY: Only gravity does work, so apply Eq. (7.4). se F ma to calculate the tension. SET P: Let y 0 at the bottom of the arc. Let point be when the string makes a 45 angle with the vertical and point be where the string is vertical. The rock moves in an arc of a circle, so it has radial acceleration a rad / v r EXECTE: (a) At the top of the swing, when the kinetic energy is zero, the potential energy (with respect to the bottom of the circular arc) is mgl( cos ), where l is the length of the string and is the angle the string

7 makes with the vertical. At the bottom of the swing, this potential energy has become kinetic energy, so mgl( cos ) mv, or v gl( cos ) (980 m/s )(080 m)( cos45 ) m/s. (b) At 45 from the vertical, the speed is zero, and there is no radial acceleration; the tension is equal to the radial component of the weight, or mg cos (0 kg)(980 m/s ) cos N. (c) At the bottom of the circle, the tension is the sum of the weight and the mass times the radial acceleration, mg mv / l mg( ( cos45 )) 9 N EVALATE: When the string passes through the vertical, the tension is greater than the weight because the acceleration is upward.