Chapter 7 Solutions. The normal force and the weight are both at 90 to the motion. Both do 0 work.

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1 Chapter 7 Solutions *7. W Fd (5000 N)(3.00 km) 5.0 MJ *7. The component of force along the direction of motion is F cos θ (35.0 N) cos N The work done by this force is W (F cos θ)d (3.7 N)(50.0 m) J 7.3 (a) W mgh ( )(9.80)(00) J J Since R mg, W air resistance J 7.4 (a) ΣF y F sin θ + n mg 0 d 0.0 m n mg F sin θ ΣF x F cos θ µ k n 0 n n F cos θ µ k f R µ k n 8.0 kg F t θ 0.0 mg F sin θ F cos θ µ k µ k mg F µ k sin θ + cos θ mg (0.500)(8.0)(9.80) F sin cos N (c) W F Fd cos θ (79.4 N)(0.0 m) cos kj f k F cos θ 74.6 N W f f k d cos θ (74.6 N)(0.0 m) cos kj 7.5 (a) W Fd cos θ (6.0 N)(.0 m) cos J and (c) The normal force and the weight are both at 90 to the motion. Both do 0 work. (d) W 3.9 J J

2 Chapter 7 Solutions 7.6 F y ma y d 5.00 m n + (70.0 N) sin N 0 n 3 N f k µ k n (3 N) 36.9 N (a) W Fd cos θ n 70.0 N sin 0.0 f k 70.0 N cos kg (70.0 N)(5.00 m) cos J W Fd cos θ (3 N)(5.00 m) cos J mg 47 N (c) W Fd cos θ (47 N)(5.00 m) cos (d) W Fd cos θ (36.9 N)(5.00 m) cos J (e) K K f K i W 39 J 85 J +44 J 7.7 W mg( y) mg(l l cos θ) (80.0 kg)(9.80 m/s )(.0 m)( cos 60.0 ) 4.70 kj l 60 y 7.8 A 5.00; B 9.00; θ 50.0 A B AB cos θ (5.00)(9.00) cos A B AB cos θ 7.00(4.00) cos ( ) A B (A x i + A y j + A z k) (B x i + B y j + B z k) A B A x B x (i i) + A x B y (i j) + A x B z (i k) + A y B x (j i) + A y B y (j j) + A y B z (j k) + A z B x (k i) + A z B y (k j) + A z B z (k k)

3 Chapter 7 Solutions 3 A B A x B x + A y B y + A z B z

4 4 Chapter 7 Solutions 7. (a) W F d F x x + F y y (6.00)(3.00) N m + (.00)(.00) N m 6.0 J θ cos F d Fd cos 6 [(6.00) + (.00) ][(3.00) + (.00) ] A B (3.00i + j k) ( i +.00j k) A B 4.00i j 6.00k C (A B) (.00j 3.00k) (4.00i j 6.00k) 0 + (.00) + (+8.0) (a) A 3.00i.00j B 4.00i 4.00j θ cos A B AB cos (3.0)(3.0).3 B 3.00i 4.00j +.00k A.00i j cos θ A B AB (0.0)(9.0) θ 56 (c) A i.00j +.00k B 3.00j k θ cos A B AB cos *7.4 We must first find the angle between the two vectors. It is: θ y Then 8 x or F v Fv cos θ (3.8 N)(0.73 m/s) cos 0.0 F 3.8 N t θ 3 F v 5.33 N m s 5.33 J s 5.33 W v 7.3 cm/s

5 Chapter 7 Solutions W i f Fdx area under curve from x i to x f (a) x i 0 x f 8.00 m W area of triangle ABC W m 6.00 N 4.0 J AC altitude, F x (N) 6 B 4 A 0 C E D x (m) x i 8.00 m x f 0.0 m W area of CDE CE altitude, W 8 0 (.00 m) ( 3.00 N) 3.00 J (c) W 0 0 W W ( 3.00).0 J *7.6 F x (8x 6) N (a) F x (N) (3, 8) 3 4 x (m) W net 7.7 W F x dx and (.00 m)(6.0 N) + (.00 m)(8.00 N).0 J W equals the area under the Force-Displacement Curve F x (N) (a) For the region 0 x 5.00 m, 3 W (3.00 N)(5.00 m) For the region 5.00 x 0.0, 7.50 J x (m) W (3.00 N)(5.00 m) 5.0 J

6 6 Chapter 7 Solutions (c) For the region 0.0 x 5.0, W (3.00 N)(5.00 m) 7.50 J (d) For the region 0 x 5.0 W ( ) J 30.0 J 7.8 W i f F ds 0 5 m (4xi + 3yj) N dxi 5 m (4 N/m) x dx + 0 (4 N/m)x / 5 m J *7.9 k F y Mg y (4.00)(9.80) N.50 0 m N/m (a) For.50 kg mass y mg k (.50)(9.80) cm Work ky Work ( N m)( m).5 J 7.0 (a) Spring constant is given by F kx k F x (30 N) (0.400 m) 575 N/m Work F avg x (30 N)(0.400 m) 46.0 J 7. Compare an initial picture of the rolling car with a final picture with both springs compressed K i + W K f 500 F (N) d (cm)

7 Chapter 7 Solutions 7 Use equation 7.. K i + k (x i x f ) + k (x i x f ) K f mv i + 0 (600 N/m)(0.500 m) + 0 (3400 N/m)(0.00 m) 0 (6000 kg) v i 00 J 68.0 J 0 v i 68 J/6000 kg 0.99 m/s 7. (a) W i f F ds W m (5000 N x N/m 5000 x N/m ) dx cos 0 W 5,000x + 0,000x 5,000x W 9.00 kj +.80 kj.80 kj 9.00 kj Similarly, W (5.0 kn)(.00 m) + (0.0 kn/m)(.00 m) (5.0 kn/m )(.00 m) 3 3 W.7 kj, larger by 9.6% J k(0.00 m) k 800 N/m and to stretch the spring to 0.00 m requires W (800)(0.00) 4.00 J.0 J Goal Solution G: We know that the force required to stretch a spring is proportional to the distance the spring is stretched, and since the work required is proportional to the force and to the distance, then W x. This means if the extension of the spring is doubled, the work will increase by a factor of 4, so that for x 0 cm, W 6 J, requiring J of additional work. O: Let s confirm our answer using Hooke s law and the definition of work.

8 8 Chapter 7 Solutions A: The linear spring force relation is given by Hooke s law: F s kx Integrating with respect to x, we find the work done by the spring is: W s x y Fs dx x y ( kx)dx x x xx k (x f x i ) However, we want the work done on the spring, which is W W s k(x f x i ) We know the work for the first 0 cm, so we can find the force constant: k W 0 0 x 0 0 (4.00 J) (0.00 m) 800 N/m Substituting for k, x i and x f, the extra work for the next step of extension is W (800 N/m) [(0.00 m) (0.00 m) ].0 J L: Our calculated answer agrees with our prediction. It is helpful to remember that the force required to stretch a spring is proportional to the distance the spring is extended, but the work is proportional to the square of the extension. 7.4 W kd k W d W k(d) kd W 3 kd 3W 7.5 (a) The radius to the mass makes angle θ with the horizontal, so its weight makes angle θ with the negative side of the x-axis, when we take the x-axis in the direction of motion tangent to the cylinder. F x ma x x F R mg t θ t θ n F mg cos θ 0 F mg cos θ

9 Chapter 7 Solutions 9 W i f F ds We use radian measure to express the next bit of displacement as ds r dθ in terms of the next bit of angle moved through: W 0 0 π/ mg cos θ Rdθ mgr sin θ π/ W mgr ( 0) mgr *7.6 [k] F x N m kg m/s m kg s 7.7 (a) K A (0.600 kg)(.00 m/s).0 J mv B K B v B K B m ()(7.50) m/s (c) W K K B K A m(v B v A ) 7.50 J.0 J 6.30 J *7.8 (a) K mv (0.300 kg)(5.0 m/s) 33.8 J K (0.300)(30.0) (0.300)(5.0) (4) 4(33.8) 35 J 7.9 v i (6.00i.00j) m/s (a) v i v i x + v iy 40.0 m/s K i mv i (3.00 kg)(40.0 m /s ) 60.0 J v 8.00i j v v v m /s K K K i m(v v i ) 3.00 (80.0) J

10 0 Chapter 7 Solutions 7.30 (a) K W (500 kg) v 5000 J v.00 m/s W F d 5000 J F(5.0 m) F 00 N 7.3 (a) K mv 0 W, so v W/m and v W/m W F d F x d F x W/d 7.3 (a) K K f K i mv f 0 W (area under curve from x 0 to x 5.00 m) v f (area) m (7.50 J) 4.00 kg.94 m/s K K f K i mv f 0 W (area under curve from x 0 to x 0.0 m) v f (area) m (.5 J) 4.00 kg 3.35 m/s (c) K K f K i mv f 0 W (area under curve from x 0 to x 5.0 m) v f (area) m (30.0 J) 4.00 kg 3.87 m/s d 5.00 m *7.33 F y ma y n 39 N 0 n 39 N n f k µ k n 0.300(39 N) 8 N (a) W F Fd cos θ (30)(5.00) cos J f k 40.0 kg F 30 N W fk f k d cos θ (8)(5.00) cos J mg 39 N

11 Chapter 7 Solutions (c) W n nd cos θ (39)(5.00) cos 90 0 (d) W g mg cos θ (39)(5.00) cos ( 90 ) 0 (e) K K f K i W mv f J 588 J J (f) v f K f m (6.0 J).76 m/s 40.0 kg 7.34 (a) K i + W K f mv f 0 + W ( kg)(780 m/s) 4.56 kj W F d cos θ J (0.70 m) cos kn (c) a v f v i x (780 m/s) 0 (0.70 m) 4 km/s (d) F ma (5 0 3 kg)(4 0 3 m/s ) 6.34 kn 7.35 (a) W g mgl cos ( θ) (0.0 kg)(9.80 m/s )(5.00 m) cos 0 68 J f k µ k n µ k mg cos θ W f lf k lµ k mg cos θ cos 80 l v i F 00 N W f (5.00 m)(0.400)(0.0)(9.80) cos J (c) W F Fl (00)(5.00) 500 J t θ (d) K W W F + W f + W g 48 J (e) K mv f mv i v f ( K) m + v i (48) (.50) 5.65 m/s

12 Chapter 7 Solutions 7.36 W K 0 L 0 d mg sin 35.0 dl 0 kx dx 0 mg sin 35.0 (L) kd d mg sin 35.0 (L) k (.0 kg)(9.80 m/s ) sin 35.0 (3.00 m) N/m 0.6 m 7.37 v i.00 m/s µ k 0.00 W K f k x 0 mv i µ k mgx mv i v i x µ k g (.00 m/s) (0.00)(9.80).04 m Goal Solution G: Since the sled s initial speed of m/s (~ 4 mph) is reasonable for a moderate kick, we might expect the sled to travel several meters before coming to rest. O: We could solve this problem using Newton s second law, but we are asked to use the workkinetic energy theorem: W K f K i, where the only work done on the sled after the kick results from the friction between the sled and ice. (The weight and normal force both act at 90 to the motion, and therefore do no work on the sled.) A: The work due to friction is W f k d where f k µ k mg. Since the final kinetic energy is zero, W K 0 K i mv i Solving for the distance d mv i µ k mg v i µ k g (.00 m) (0.00)(9.80 m/s ).04 m L: The distance agrees with the prediction. It is interesting that the distance does not depend on the mass and is proportional to the square of the initial velocity. This means that a small car and a massive truck should be able to stop within the same distance if they both skid to a stop from the same initial speed. Also, doubling the speed requires 4 times as much stopping distance, which is consistent with advice given by transportation safety officers who suggest at least a second gap between vehicles (as opposed to a fixed distance of 00 feet).

13 Chapter 7 Solutions (a) v f 0.0c 0 ( m/s) m/s K f mv f ( kg)( m/s) J K i + Fd cos θ K f 0 + F(0.360 m) cos N m F N (c) a F m N kg.5 03 m/s (d) x f x i (v i+ v f ) t t (x f x i ) (v i + v f ) (0.360 m) ( m/s) s 7.39 (a) W K fd cos θ mv f mv i f( m) cos 80 0 ( kg)(600 m/s) f N t d v m [ m/s]/ s 7.40 W K m gh m gh (m + m ) v f 0 v f (m m )gh m + m ( )(9.80)(0.400) m s v f.57 m/s.5 m/s 7.4 (a) W s kx i kx f (500)( ) J W mv f mv i mv f 0 so v f ( W) m (0.65).00 m/s 0.79 m/s

14 4 Chapter 7 Solutions W W s + W f 0.65 J + ( µ k mgd) 0.65 J (0.350)(.00)(9.80)( ) J 0.8 J v f ( W) m (0.8).00 m/s 0.53 m/s *7.4 A 300-kg car speeds up from rest to 55.0 mi/h 4.6 m/s in 5.0 s. The output work of the engine becomes its final kinetic energy, (300 kg)(4.6 m/s) 390 kj with power J 5.0 s ~ 0 4 W, around 30 horsepower Power W t mgh t (700 N)(0.0 m) 8.00 s 875 W 7.44 Efficiency e useful energy output/total energy input. The force required to lift n bundles of shingles is their weight, nmg. e n mgh cos 0 Pt n ept mgh (0.700)(746 W)(700 s) (70.0 kg)(9.80 m/s )(8.00 m) kg m s 3 W 685 bundles 7.45 P a f a v f a P a v N *7.46 (a) W K, but K 0 because he moves at constant speed. The skier rises a vertical distance of (60.0 m) sin m. Thus, W in W g (70.0 kg)g(30.0 m) J 0.6 kj The time to travel 60.0 m at a constant speed of.00 m/s is 30.0 s. Thus, P input W t J 30.0 s 686 W 0.99 hp

15 Chapter 7 Solutions (a) The distance moved upward in the first 3.00 s is y v t m/s (3.00 s).63 m W mv f mv i + mgy f mgy i mv f mv i + mg( y) W (650 kg)(.75 m/s) 0 + (650 kg)g(.63 m) J Also, W P t so P W t J 3.00 s W 7.9 hp When moving upward at constant speed (v.75 m/s), the applied force equals the weight (650 kg)(9.80 m/s ) N. Therefore, P Fv ( N)(.75 m/s). 0 4 W 4.9 hp *7.48 energy power time For the 8.0 W bulb: Energy used (8.0 W)( h) 80 kilowatt hrs total cost $ (80 kwh)($0.080/kwh) $39.40 For the 00 W bulb: Energy used (00 W)( h) kilowatt hrs # bulb used h 750 h/bulb 3.3 total cost 3.3($0.40) + ( kwh)($0.080/kwh) $85.60 Savings with energy-efficient bulb $85.60 $39.40 $46.

16 6 Chapter 7 Solutions mv f mv i 7.49 (a) fuel needed useful energy per gallon mv f 0 eff. (energy content of fuel) 73.8 (900 kg)(4.6 m/s) (0.50)( J/gal).35 0 gal (c) power gal 38.0 mi 55.0 mi.00 h.00 h J 3600 s (0.50) 8.08 kw gal 7.50 At a speed of 6.8 m/s (60.0 mph), the car described in Table 7. delivers a power of P 8.3 kw to the wheels. If an additional load of 350 kg is added to the car, a larger output power of P P + (power input to move 350 kg at speed v) will be required. The additional power output needed to move 350 kg at speed v is: P out ( f)v (µ r mg)v Assuming a coefficient of rolling friction of µ r 0.060, the power output now needed from the engine is P P + (0.060)(350 kg)(9.80 m/s )(6.8 m/s) 8.3 kw +.47 kw With the assumption of constant efficiency of the engine, the input power must increase by the same factor as the output power. Thus, the fuel economy must decrease by this factor: (fuel economy) P P (fuel economy) km L or (fuel economy) 5.9 km L

17 Chapter 7 Solutions When the car of Table 7. is traveling at 6.8 m/s (60.0 mph), the engine delivers a power of P 8.3 kw to the wheels. When the air conditioner is turned on, an additional output power of P.54 kw is needed. The total power output now required is P P + P 8.3 kw +.54 kw Assuming a constant efficiency of the engine, the fuel economy must decrease by the same factor as the power output increases. The expected fuel economy with the air conditioner on is therefore (fuel economy) P P (fuel economy) km L or (fuel economy) 5.90 km L 7.5 (a) K (v/c) mc (0.995) ( )( ) K J Classically, K mv ( kg) [(0.995)( m/s)] J This differs from the relativistic result by % error J J J 00% 94.5% 7.53 W K f K i (v f /c) mc (v i /c) mc or (a) W W (v f /c) mc (v i /c) (0.750) ( kg)( m/s) (0.500) W J W (0.995) ( kg)( m/s) (0.500) W J

18 8 Chapter 7 Solutions Goal Solution G: Since particle accelerators have typical maximum energies on the order of GeV (ev J), we could expect the work required to be ~0 0 J. O: The work-energy theorem is W K f K i which for relativistic speeds (v ~ c) is: W mc v f/c mc v i/c A: (a) W (0.750) ( kg)( m/s) (0.500) ( J) W ( )( J) J E (0.995) ( J) (.55 )( J) W ( )( J) J L: Even though these energies may seem like small numbers, we must remember that the proton has very small mass, so these input energies are comparable to the rest mass energy of the proton (938 MeV J). To produce a speed higher by 33%, the answer to part is 5 times larger than the answer to part (a). Even with arbitrarily large accelerating energies, the particle will never reach or exceed the speed of light. This is a consequence of special relativity, which will be examined more closely in a later chapter. *7.54 (a) Using the classical equation, K mv (78.0 kg)( m/s) J Using the relativistic equation, K (v/c) mc ( / ) (78.0 kg)( m/s) K J

19 Chapter 7 Solutions 9 When (v/c) <<, the binomial series expansion gives [ (v/c) ] / + (v/c) Thus, [ (v/c) ] / (v/c) and the relativistic expression for kinetic energy becomes K (v/c) mc mv. That is, in the limit of speeds much smaller than the speed of light, the relativistic and classical expressions yield the same results. *7.55 At start, v (40.0 m/s) cos 30.0 i + (40.0 m/s) sin 30.0 j At apex, v (40.0 m/s) cos 30.0 i + 0j 34.6i m/s and K mv (0.50 kg)(34.6 m/s) 90.0 J *7.56 Concentration of Energy output (0.600 J/kg step)(60.0 kg) step.50 m 4.0 J m F 4.0 J m N m J 4.0 N P Fv 70.0 W (4.0 N)v v.9 m/s 7.57 The work-kinetic energy theorem is K i + W K f The total work is equal to the work by the constant total force: mv i + (ΣF) (r r i ) mv f mv i + ma (r r i ) mv f v i + a (r r i ) v f

20 0 Chapter 7 Solutions 7.58 (a) A i (A)() cos α. But also, A i A x. Thus, (A)() cos α A x or cos α A x A Similarly, cos β A y A and cos γ A z A where A A x + A y + A z cos α + cos β + cos γ A x + A y + A z A A A A A 7.59 (a) x t +.00t 3 therefore, v dx dt t K mv (4.00)( t ) ( t + 7.0t 4 ) J a dv dt (.0t) m/s F ma 4.00(.0t) (48.0t) N (c) P Fv (48.0t)( t ) (48.0t + 88t 3 ) W (d) W 0.00 P dt 0.00 (48.0t + 88t 3 ) dt 50 J *7.60 (a) The work done by the traveler is mgh s N where N is the number of steps he climbs during the ride. N (time on escalator)(n) where (time on escalator) h vertical velocity of person, and vertical velocity of person v + nh s

21 Chapter 7 Solutions Then, N nh v + nh s and the work done by the person becomes W person mgnhh s v + nh s The work done by the escalator is W e (power)(time) [(force exerted)(speed)](time) mgvt h where t v + nh s as above. Thus, W e mgvh v + nh s As a check, the total work done on the person s body must add up to mgh, the work an elevator would do in lifting him. It does add up as follows: W W person + W e mgnhh s v + nh s + mgvh v + nh s mgh(nh s + v) v + nh s mgh 7.6 W x i x f F dx 0 x f ( kx + βx 3 ) dx W kx + βx4 4 xf 0 kx f + βx4 f 4 W ( 0.0 N/m)(0.00 m) + (00 N/m3 )(0.00 m) 4 4 W J J J *7.6 F x ma x kx ma k ma x ( kg)0.800(9.80 m/s ) m 7.37 N/m n a F s m m mg

22 Chapter 7 Solutions 7.63 Consider the work done on the pile driver from the time it starts from rest until it comes to rest at the end of the fall. W K W gravity + W beam mv f mv i so (mg)(h + d) cos 0 + (F )(d) cos Thus, F (mg)(h + d) d (00 kg)(9.80 m/s )(5. m) 0.0 m N Goal Solution G: Anyone who has hit their thumb with a hammer knows that the resulting force is greater than just the weight of the hammer, so we should also expect the force of the pile driver to be greater than its weight: F > mg ~0 kn. The force on the pile driver will be directed upwards. O: The average force stopping the driver can be found from the work that results from the gravitational force starting its motion. The initial and final kinetic energies are zero. A: Choose the initial point when the mass is elevated and the final point when it comes to rest again 5. m below. Two forces do work on the pile driver: gravity and the normal force exerted by the beam on the pile driver. W net K f K i so that mgs w cos 0 +ns n cos 80 0 where m 00 kg, s w 5. m, and s n 0.0 m. In this situation, the weight vector is in the direction of motion and the beam exerts a force on the pile driver opposite the direction of motion. (00 kg) (9.80 m/s ) (5. m) n (0.0 m) 0 Solve for n. n J 0.0 m 878 kn (upwards) L: The normal force is larger than 0 kn as we expected, and is actually about 43 times greater than the weight of the pile driver, which is why this machine is so effective.

23 Chapter 7 Solutions 3 Additional Calculation: Show that the work done by gravity on an object can be represented by mgh, where h is the vertical height that the object falls. Apply your results to the problem above. By the figure, where d is the path of the object, and h is the height that the object falls, d θ d y h h d y d cos θ Since F mg, mgh Fd cos θ F d In this problem, mgh n(d n ), or (00 kg)(9.80 m/s )(5. m) n(0.0 m) and n 878 kn 7.64 Let b represent the proportionality constant of air drag f a to speed: f a bv Let f r represent the other frictional forces. Take x-axis along each roadway. For the gentle hill F x ma x bv f r + mg sin.00 0 b(4.00 m/s) f r N 0 For the steeper hill b(8.00 m/s) f r N 0 Subtracting, b(4.00 m/s) 5.6 N b 6.40 N s/m and then f r N. Now at 3.00 m/s the vehicle must pull her with force bv + f r (6.40 N s/m)(3.00 m/s) N 9. N and with power P F v 9. N(3.00 m/s) cos W

24 4 Chapter 7 Solutions 7.65 (a) P Fv F(v i + at) F 0 + F m t m F t P (0.0 N) (3.00 s) 40 W 5.00 kg 7.66 (a) The new length of each spring is x + L, so its extension is x + L L and the force it exerts is k ( x + L L) toward its fixed end. The y components of the two spring forces add to zero. Their x components add to L L A k m x F ik ( x + L L)x/ x + L k F kxi ( L/ x + L ) (top view) f W i F x dx W A 0 kx ( L/ x + L )dx W k A 0 x dx + kl A 0 (x + L ) / x dx W k x 0 A + kl (x + L ) / (/) 0 A W 0 + ka + kl kl A + L W kl + ka kl A + L 7.67 (a) F (5.0 N)(cos 35.0 i + sin 35.0 j) (0.5i + 4.3j) N F (4.0 N)(cos 50 i + sin 50 j) ( 36.4 i +.0 j) N F F + F ( 5.9i j) N (c) a F m ( 3.8i j) m/s (d) v v i + at (4.00i +.50j) m/s + ( 3.8i j)(m/s )(3.00 s) v ( 5.54i + 3.7j) m/s

25 Chapter 7 Solutions 5 (e) r r i + v i t + at r 0 + (4.00i +.50j)(m/s)(3.00 s) + ( 3.8i j)(m/s )(3.00 s) d r (.30i j) m (f) K f mv f (5.00 kg) [(5.54) + (3.7) ](m/s).48 kj (g) K f mv i + F d (5.00 kg) [(4.00) + (.50) ](m/s) + [( 5.9 N)(.30 m) + (35.3 N)(39.3 m)] 55.6 J + 46 J.48 kj 7.68 (a) F (N) L (mm) F (N) L (mm) F (N) L (mm) (c) A straight line fits the first eight points, and the origin. By least-square fitting, its slope is 0.5 N/mm ± % 5 N/m ± %. In F kx, the spring constant is k F/x, the same as the slope of the F-versus-x graph. F kx (5 N/m)(0.05 m) 3. N

26 6 Chapter 7 Solutions 7.69 (a) W K W s + W g 0 ( N/m) (0.00 m) (0.00 kg)(9.80)(sin 60.0 )x 0 x 4. m W K W s + W g + W f 0 ( N/m) (0.00) [(0.00)(9.80)(sin 60.0 ) + (0.00)(9.80)(0.400)(cos 60.0 )]x 0 x 3.35 m *7.70 (a) W K m(v f v i ) (0.400 kg) [(6.00) (8.00) ] (m/s) 5.60 J W fd cos 80 µ k mg(πr) 5.60 J µ k (0.400 kg)(9.80 m/s )π(.50 m) Thus, µ k 0.5 (c) After N revolutions, the mass comes to rest and K f 0. Thus, W K 0 K i mv i or µ k mg [N(πr)] mv i This gives mv i (8.00 m/s) N µ k mg(πr) (0.5)(9.80 m/s.8 rev )π(.50 m)

27 Chapter 7 Solutions N (5.00 cm)( m) cm (0.00 kg)(9.80 m/s )( m) sin (0.00 kg) v 0.50 J J + ( kg)v v m/s If positive F represents an outward force, (same direction as r), then f W i F ds r i r f (F0 σ 3 r 3 F 0 σ 7 r 7 ) dr W +F 0σ 3 r ( ) F 0σ 7 r 6 ( 6) rf r i W F 0σ 3 (r f r i ) 6 + F 0σ 7 (r 6 f r 6 i ) 6 F 0σ 7 6 [r 6 f r 6 i ] F 0σ 3 6 [r f r i ] W [r 6 f r 6 i ] [r f r i ] W [ ] [ ] 0 0 W.49 0 J +. 0 J.37 0 J 7.73 (a) W K m gh µm gh (m + m )(v v i) v gh(m µm ) (m + m ) (9.80)(0.0)[0.400 (0.00)(0.50)] ( ) 4.5 m/s

28 8 Chapter 7 Solutions W f + W g K 0 µ( m + m )gh + m gh 0 µ( m + m ) m m m µ m kg 0.00 (c) W f + W g K 0 µm gh + (m m )gh kg.75 kg m m µm kg (0.00)(0.50 kg) kg 7.74 P t W K ( m)v v t The density is ρ m vol m A x Substituting this into the first equation and solving for P, since A v x t v for a constant speed, we get P ρav3 Also, since P Fv, F ρav 7.75 We evaluate dx x x by calculating 375(0.00) (.8) (.8) + 375(0.00) (.9) (.9) (0.00) (3.6) (3.6) and 375(0.00) (.9) (.9) + 375(0.00) (3.0) (3.0) (0.00) (3.7) (3.7) 0.79 The answer must be between these two values. We may find it more precisely by using a value for x smaller than Thus, we find the integral to be N m.

29 Chapter 7 Solutions 9 *7.76 (a) The suggested equation P t bwd implies all of the following cases: () P t b w (d) () P t b w d (3) P t bw d and (4) P t b w d These are all of the proportionalities Aristotle lists. v constant n d f k µ k n F w For one example, consider a horizontal force F pushing an object of weight w at constant velocity across a horizontal floor with which the object has coefficient of friction µ k. F ma implies that: +n w 0 and F µ k n 0 so that F µ k w As the object moves a distance d, the agent exerting the force does work W Fd cos θ Fd cos 0 µ k wd and puts out power P W/t This yields the equation P t µ k wd which represents Aristotle s theory with b µ k. Our theory is more general than Aristotle s. Ours can also describe accelerated motion.

The negative root tells how high the mass will rebound if it is instantly glued to the spring. We want

The negative root tells how high the mass will rebound if it is instantly glued to the spring. We want 8.38 (a) The mass moves down distance.0 m + x. Choose y = 0 at its lower point. K i + U gi + U si + E = K f + U gf + U sf 0 + mgy i + 0 + 0 = 0 + 0 + kx (.50 kg)9.80 m/s (.0 m + x) = (30 N/m) x 0 = (60