PH201 Chapter 7 Solutions

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1 PH0 Chapter 7 Solutions 7.6. Set Up: Use W F s ( F cos ) s Calculate the work done by each force. In each case, identify the angle In part (d), the net work is the algebraic sum of the work done by each force. Since the force exerted by the rope and the displacement are in the same direction, 0 and W rope (70 N)(cos0 )(50 m) 74 J (b) Gravity is downward and the displacement is at 0 0 above the horizontal, so W grav (8 0 N)(cos0 )(50 m) J (c) The normal force n is perpendicular to the surface of the ramp while the displacement is parallel to the surface of the ramp, so 90 and W 0 n (d) Wnet Wrope Wgrav W n 74 J J 0 4 J (e) Now and W rope (70 N)(cos0 0 )(50 m) 5 J Reflect: In part (b), gravity does negative work since the gravity force acts downward and the carton moves upward. Less work is done by the rope in part (e), but the net work is still positive. *7.5. Set Up: Use K m to relate and K. Let and K be speed and kinetic energy when 00 m/s K K / and K / K / so K /K (0 0 m/s) / 707 m/s (b) Since /, we obtain K K ( / ) K (/4) This means that the decrease in K is K/ Set Up: Since the meteor comes to rest the energy it delivers to the ground equals its original kinetic energy: 4 K mv. We know that v km/s 0 m/s K ( 4 0 kg)( 0 m/s) 0 0 J J 7 7 h day (b) It would take a typical coal-fired power plant 0 s, which is equal to (0 s) J/s 600 s 4 h days, to produce this much energy. Reflect: Part of the energy transferred to the ground lifts soil and rocks into the air and creates a large crater. *7.. Set Up: From the work-energy relation, W Wgrav Krock or F s Kf Ki As the rock rises, the gravitational force, F mg, does work on the rock. Since this force acts in the direction opposite to the motion and displacement, s, the work is negative. Applying F s Kf Ki we obtain: f i mgh m m Dividing by m and solving for i, i f gh Substituting h 5 0 m and f 5 0 m/s, i (50 m/s) (980 m/s )(5 0 m) 0 m/s (b) Solve the same work-energy relation for h. At the maximum height f 0

2 f i i f (0 m/s) (00 m/s) mgh m m h 468 m g (980 m/s ) Reflect: Note that the weight of 0 N was never used in the calculations because both gravitational potential and kinetic energy are proportional to mass, m. Thus any object, that attains 5 0 m/s at a height of 5.0 m, must have an initial velocity of 0 m/s As the rock moves upward gravity does negative work and this reduces the kinetic energy of the rock Set Up: Apply W tot K K, where K 0. The normal force does no work. The work W done by gravity is W mgh, where h Lsin is the vertical distance the block has dropped when it has traveled a distance L down the incline and is the angle the plane makes with the horizontal. Solve: The work-energy theorem gives v (980 m/ s )(075 m)sin69 97 m/ s K W v gh glsin. Using the given numbers, m m Reflect: The final speed of the block is the same as if it had been dropped from a height h Set Up: Use Fon spring kx in parts (a) and (c) and W on spring kx in part (b). Fon spring 50 N k 4 0 N/m x 09 m 070 m (b) x 0 0 m so on spring W ( 4 0 N/m)(0 0 m) 0 76 J (c) Since Fon spring kx, doubling the force results in doubling the length of stretch giving x 0044 m The new length will therefore be 70 cm 44 cm 4 cm *7.5. Set Up: Use Fon tendon (b), also apply U el kx In part (a), F on tendon equals mg, the weight of the object suspended from it. In part kx to calculate the stored energy. Fon tendon (050 kg)(980 m/s ) k 99 N/m x 00 m Fon tendon 8 N (b) x 069 m 69 cm; el k 99 N/m U (99 N/m)(0 69 m) 47 8 J Reflect: The 50 g object has a weight of.45 N. The 8 N force is much larger than this and stretches the tendon a much greater distance. *7.9. Set Up: The change in gravitational potential energy is Ugrav mg( yf yi ), while the increase in kinetic energy, for a zero initial velocity, is energies developed. f The food energy equals mg( yf yi), so K m Set the food energy, expressed in joules, equal to the mechanical

3 (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): y (00)(90 m) 80 m Set Up: Set 0% of the food energy (in joules) equal to the total work you do in raising the weight N times. Solve: For each arm raise, you do work equal to W mgh where h 5 cm 486 J 00(50 food cal) N( mgh) food cal (00)(50)(486 J) N 7 0 (50 kg)(980 m/s )(05 m) Reflect: It is not reasonable to do this many repetitions in a single exercise session Set Up: In part (a), substitute Ki Kf 0 into Kf Uf Ki Ui to obtain U el,f U el,i. Also let subscript refer to the stronger spring and to the weaker spring. Then which results in m max k x kx k x N/cm x x k/ k (40 cm) (40 cm)( ) 5 7 cm 6 N/cm (b) k 00 N/m x (40 0 m) 8 m/s m 5 kg max and x x k/ k In part (b), apply Uel,i Kf Set Up: Apply Emech K U with Ui Uf since the otter returns to the same height. Thus, mech,i mech,f i f i f E E K K m( ) Solve: E mech,i E mech,f ( 4 kg)[(5 75 m/s) ( 75 m/s) ] 08 J Reflect: Part of the initial mechanical energy was converted to thermal energy by the negative work done by friction Set Up: Use the work-energy theorem, Eq. (7.6). The target variable k will be a factor in the work done by friction. Let point be where the block is initially released and let point be where the block finally stops, as shown in the figure below. Thus, we have K U W other K U. Work is done on the block by the spring and by friction, so Wother Wf and U Uel Solve: We have KK 0,,el U U kx (00 N/m)(0 00 m) 00 J, and U U,el 0 (since after the block leaves the spring has given up all its stored energy). Finally, we have Wother Wf ( fk cos ) s mg(cos ) s mgs, since 80 (The friction force is directed opposite to the displacement and does negative k k

4 work). Putting all this into K U Wother K U gives U,el W f 0, so k mgs U,el. Solving for k we obtain U,el 00 J k 04 mgs (050 kg)(980 m/s )( 00 m) Reflect: U,el W f 0 says that the potential energy originally stored in the spring is taken out of the system by the negative work done by friction. *7.65. Set Up: Since the electricity cost rate is given in cents per kwh, each power must be multiplied by the corresponding time usage in determining the cost: cost Pt (cost/ Pt) (kwh)[cents/(kwh)] cos t 746 W kw 75. cos t Pt (0 0 hp)(8 00 hr) 49. $49 Pt hp 000 W kwh cost 40 hr kw 75.. (b) cos t Pt (75 W) $0 per day Pt day 000 W kwh day Reflect: Note that the cost of running the motor for an entire day is ($49) $ ; thus, the cost of running the motor is ($ )/($0 ) 0 00 times the cost of running the light. *7.69. Set Up: From Problem 7.68, the U.S. consumes 0 W of power per year. The area required in part (a) may then be calculated as PU S radiation A( )/(P /m ) However, for an array efficiency of 5%, the radiative power rate must be multiplied by 0.5 to account for power losses. For part (b), since the array is square, simply L A A L and the length of a side is 0 W km U S radiation 6 A ( P )/(P /m ) 0 km 05( 0 0 W/m ) 0 m (b) Lmetric A 0 km 6 km L English mi (6 km) mi 609 km Reflect: An array of this size is approximately equivalent to the state of Rhode Island and thus feasible. *7.75. Set Up: For part (a), the work performed each day is W Wgrav mgh where h 6 m The mass of the blood is calculated using the density, mv / 050 kg/m, and the known volume of V 7500 L The power output is then found from P W grav / t The work performed in a single day is: kg m m 5 W Wgrav mgh Vgh L 9 80 ( 6 m) 6 0 J 000 L m s W 5 grav (6 0 J/day) (b) P 46 J/s 46 W t (4 hr/day)(600 s/hr) (c) Reflect: The heart actually puts out more power than the value calculated because two forms of energy were not considered; the heart also provides kinetic energy to the blood and generates thermal energy by working against the friction of the vein and vessel walls.

5 7.80. Set Up: Apply Kf Uf Ki Ui to the system of the two climbers for which Ki 0 For the 60.0 kg person, who experiences a gravitational potential energy change, let yf 0 so yi 50 m Both climbers move with the same speed Substituting into the conservation equation, m f m f mgyi, where m 75 0 kg and m 600 kg Solve: mgyi (600 kg)(980 m/s )( 50 m) f 6 m/s m m 750 kg 600 kg *7.97. Set Up: For parts (a) and (c), use the energy relation W P t to find the energy in joules and then convert the respective results to food calorie units. 6 Energy W (80 J/s)(600 s/h)(4 h/day) 69 0 J/day 6 food cal W (69 0 J/day) 6 0 food calories/day 486 J (b) The body s metabolic processes convert food energy to thermal energy. (c) Energy W [(80 J/s)(600 s/h)(6 hr) (00 J/s)(600 s/h)(8 h)] J 48 0 food calories Set Up: Apply conservation of energy: Kf Uf Ki Ui Let yi 0, so yf h, the maximum height. At this maximum height, f, y 0 and f, x i, x thus f i, x (5 m/s)(cos60 0 ) 75 m/s Substituting into conservation of energy gives Solve:. m i mgh m(7 5 m/s) i (7 5 m/s) (5 m/s) (7 5 m/s) h 86 m g (980 m/s ) 7.0. Set Up: The mass is the density times the volume V. For a sphere, V quantities to SI units. 4 r Use m (5 0 kg/m ) r (5 0 kg/m ) (5 0 m) 8 0 kg K m and convert all (b) 5 m K ( 8 0 kg)( 0 m/s) 0 J (c) The time required for the human race to use 0 J is 0 J t 0 years J/y

Reflect: When the force and displacement are in opposite directions, the work done is negative.

Reflect: When the force and displacement are in opposite directions, the work done is negative. WORK AND ENERGY 7 Answers to Multiple Choice Problems. B. D 3. A 4. C 5. C 6. C 7. A, D 8. B, C, E 9. A, D, E 0. B, D. B, D. B, D 3. B, D 4. A 5. C olutions to Problems 7.. et Up: Assume the fisherman