Chapte 6 16. (a) In this situation, we take f s to point uphill and to be equal to its maximum value, in which case f s, max = μsf applies, whee μ s = 0.5. pplying ewton s second law to the block of mass m = W/g = 8. kg, in the x and y diections, poduces Fmin 1 mgsinθ + fs, max = ma = 0 F mgcos θ = 0 which (with θ = 0 ) leads to ( ) Fmin 1 mg sinθ + μs cosθ = 8.6. (b) ow we take f s to point downhill and to be equal to its maximum value, in which case f s, max = μ s F applies, whee μ s = 0.5. pplying ewton s second law to the block of mass m = W/g = 8. kg, in the x and y diections, poduces Fmin = mgsin θ fs, max = ma = 0 F mgcos θ = 0 which (with θ = 0 ) leads to ( ) Fmin = mg sin θ + μs cos θ = 46. value slightly lage than the exact esult of this calculation is equied to make it acceleate uphill, but since we quote ou esults hee to two significant figues, 46 is a good enough answe. (c) Finally, we ae dealing with kinetic fiction (pointing downhill), so that 0= F mgsinθ fk = ma 0= F mg cosθ along with f k = μ k F (whee μ k = 0.15) bings us to b g 39. F = mg sin θ + μk cos θ = 17. If the block is sliding then we compute the kinetic fiction fom Eq. 6-; if it is not sliding, then we detemine the extent of static fiction fom applying ewton s law, with zeo acceleation, to the x axis (which is paallel to the incline suface). The question of whethe o not it is sliding is theefoe cucial, and depends on the maximum static 17
18 CHPTER 6 fiction foce, as calculated fom Eq. 6-1. The foces ae esolved in the incline plane coodinate system in Figue 6-5 in the textbook. The acceleation, if thee is any, is along the x axis, and we ae taking uphill as +x. The net foce along the y axis, then, is cetainly zeo, which povides the following elationship: F y = 0 F = Wcosθ whee W = mg = 45 is the weight of the block, and θ = 15 is the incline angle. Thus, F = 43.5, which implies that the maximum static fiction foce should be (a) Fo P = f s,max = (0.50) (43.5 ) = 1.7. ( 5.0 )i ˆ, ewton s second law, applied to the x axis becomes f P mgsin θ = ma. Hee we ae assuming f is pointing uphill, as shown in Figue 6-5, and if it tuns out that it points downhill (which is a possibility), then the esult fo f s will be negative. If f = f s then a = 0, we obtain o f s = ˆ (17 )i f s = P + mg sinθ = 5.0 + (43.5 )sin15 =17,. This is clealy allowed since f s is less than f s, max. (b) Fo P = ( 8.0 )i ˆ, we obtain (fom the same equation) f (0 )i ˆ s =, which is still allowed since it is less than f s, max. (c) ut fo P = ( 15 )i ˆ, we obtain (fom the same equation) f s = 7, which is not allowed since it is lage than f s, max. Thus, we conclude that it is the kinetic fiction instead of the static fiction that is elevant in this case. The esult is f = μ F ˆi = (0.34)(43.5 ) ˆi = (15 ) ˆi. k k 18. (a) We apply ewton s second law to the downhill diection: mg sinθ f = ma, whee, using Eq. 6-11, f = f k = μ k F = μ k mg cosθ. Thus, with μ k = 0.600, we have a = gsinθ μ k cosθ = 3.7 m/s
19 which means, since we have chosen the positive diection in the diection of motion (down the slope) then the acceleation vecto points uphill ; it is deceleating. With v 0 = 18.0 m/s and Δx = d = 4.0 m, Eq. -16 leads to v v ad = 0 + = 1.1 m/s. (b) In this case, we find a = +1.1 m/s, and the speed (when impact occus) is 19.4 m/s. 19. (a) The fee-body diagam fo the block is shown below. F is the applied foce, F the nomal foce of the wall on the block, f is the foce of fiction, and mg is the foce of gavity. To detemine if the block falls, we find the magnitude f of the foce of fiction equied to hold it without acceleating and also find the nomal foce of the wall on the block. We compae f and μ s F. If f < μ s F, the block does not slide on the wall but if f > μ s F, the block does slide. The hoizontal component of ewton s second law is F F = 0, so F = F = 1 and is μ s F = (0.60)(1 ) = 7.. The vetical component is f mg = 0, so f = mg = 5.0. Since f < μ s F the block does not slide. (b) Since the block does not move f = 5.0 and F = 1. The foce of the wall on the block is F = F ˆi + f ˆj = 1 ˆi + 5.0 ˆj whee the axes ae as shown on Fig. 6-6 of the text. w ( ) ( ) 7. Fist, we check to see if the bodies stat to move. We assume they emain at est and compute the foce of (static) fiction which holds them thee, and compae its magnitude with the maximum value μ s F. The fee-body diagams ae shown below.
0 CHPTER 6 T is the magnitude of the tension foce of the sting, f is the magnitude of the foce of fiction on body, F is the magnitude of the nomal foce of the plane on body, mg is the foce of gavity on body (with magnitude W = 10 ), and m g is the foce of gavity on body (with magnitude W = 3 ). θ = 40 is the angle of incline. We ae told the diection of f but we assume it is downhill. If we obtain a negative esult fo f, then we know the foce is actually up the plane. (a) Fo we take the +x to be uphill and +y to be in the diection of the nomal foce. The x and y components of ewton s second law become T f W sin θ = 0 F W cos θ = 0. Taking the positive diection to be downwad fo body, ewton s second law leads to W T = 0. Solving these thee equations leads to f = W W sinθ = 3 (10 ) sin 40 = 34 (indicating that the foce of fiction is uphill) and to F = W cos θ = (10 ) cos 40 = 78 which means that f s,max = μ s F = (0.56) (78 ) = 44. Since the magnitude f of the foce of fiction that holds the bodies motionless is less than f s,max the bodies emain at est. The acceleation is zeo. (b) Since is moving up the incline, the foce of fiction is downhill with magnitude f k = μ k F. ewton s second law, using the same coodinates as in pat (a), leads to
1 T fk W sin θ = ma F W cos θ = 0 W T = m a fo the two bodies. We solve fo the acceleation: ( ) ( )( ) W sin cos 3 10 sin 40 0.5 10 cos 40 W θ μkw θ a = = m + m ( 3+10) ( 9.8 m s ) = 3.9 m s. The acceleation is down the plane, i.e., a = ( 3.9 m/s )i ˆ, which is to say (since the initial velocity was uphill) that the objects ae slowing down. We note that m = W/g has been used to calculate the masses in the calculation above. (c) ow body is initially moving down the plane, so the foce of fiction is uphill with magnitude fk = μkf. The foce equations become which we solve to obtain 1.0 m s. T + fk W sin θ = ma F W cos θ = 0 W T = m a ( ) ( )( ) W sin cos 3 10 sin 40 0.5 10 cos 40 W θ+ μkw θ + a = = m + m = ( 3+10) ( 9.8 m s ) The acceleation is again downhill the plane, i.e., objects ae speeding up. a = ( 1.0 m/s )i ˆ. In this case, the Chapte 7 17. We use F to denote the upwad foce exeted by the cable on the astonaut. The foce of the cable is upwad and the foce of gavity is mg downwad. Futhemoe, the acceleation of the astonaut is a = g/10 upwad. ccoding to ewton s second law, the foce is given by
CHPTER 6 11 F mg = ma F = m( g+ a) = mg, 10 in the same diection as the displacement. On the othe hand, the foce of gavity has magnitude Fg = mg and is opposite in diection to the displacement. (a) Since the foce of the cable F and the displacement d ae in the same diection, the wok done by F is W F 11mgd 11 (7 kg)(9.8 m/s )(15 m) 4 4 = Fd = = = 1.164 10 J 1. 10 J. 10 10 (b) Using Eq. 7-7, the wok done by gavity is W = F d = mgd = = g g 4 4 (7 kg)(9.8 m/s )(15 m) 1.058 10 J 1.1 10 J (c) The total wok done is the sum of the two woks: W = W + W = =. 4 4 3 3 net F g 1.164 10 J 1.058 10 J 1.06 10 J 1.1 10 J Since the astonaut stated fom est, the wok-kinetic enegy theoem tells us that this is he final kinetic enegy. (d) Since K = 1 mv, he final speed is 3 K (1.06 10 J) v = = = 5.4 m/s. m 7 kg ote: Fo a geneal upwad acceleation a, the net wok done is Wnet = WF + Wg = Fd Fgd = m( g+ a) d mgd = mad. Since W =Δ K = mv by wok-kinetic enegy theoem, the speed of the astonaut net / would be v= ad, which is independent of the mass of the astonaut. 19. Eq. 7-15 applies, but the woding of the poblem suggests that it is only necessay to examine the contibution fom the ope (which would be the W a tem in Eq. 7-15): W a = (50 )(0.50 m) = 5 J (the minus sign aises fom the fact that the pull fom the ope is anti-paallel to the diection of motion of the block). Thus, the kinetic enegy would have been 5 J geate if the ope had not been attached (given the same displacement).
3 3. The fact that the applied foce F a causes the box to move up a fictionless amp at a constant speed implies that thee is no net change in the kinetic enegy: Δ K = 0. Thus, the wok done by F a must be equal to the negative wok done by gavity: Wa = Wg. Since the box is displaced vetically upwad by h = 0.150 m, we have Wa =+ mgh= = (3.00 kg)(9.80 m/s )(0.150 m) 4.41 J 4. (a) Using notation common to many vecto capable calculatos, we have (fom Eq. 7-8) W = dot([0.0,0] + [0, (3.00)(9.8)], [0.500 30.0º]) = +1.31 J. (b) Eq. 7-10 (along with Eq. 7-1) then leads to v = (1.31 J)/(3.00 kg) = 0.935 m/s.