Forces and the Laws of Motion Problem C

AME DATE CLASS Forces and the Laws of Motion Problem C EWTO S SECOD LAW PROBLEM SOLUTIO A 1.5 ball has an acceleration of 9.0 m/s to the left. What is the net force acting on the ball? Given: m = 1.5 a = 9.0 m/s to the left Unknown: F =? Use ewton s second law, and solve for F. ΣF = ma Because there is only one force, ΣF = F F = (1.5 )(9.0 m/s ) = 14 F = 14 to the left ADDITIOAL PRACTICE 1. David Purley, a racing driver, survived deceleration from 173 km/h to 0 km/h over a distance of 0.660 m when his car crashed. Assume that Purley s mass is 70.0. What is the average force acting on him during the crash? Compare this force to Purley s weight. (Hint: Calculate the average acceleration first.). A giant crane in Washington, D. C. was tested by lifting a.3 6 load. a. Find the magnitude of the force needed to lift the load with a net acceleration of 0 m/s. b. If the same force is applied to pull the load up a smooth slope that makes a 30.0 angle with the horizontal, what would be the acceleration? 3. When the click beetle jumps in the air, its acceleration upward can be as large as 400.0 times the acceleration due to gravity. (An acceleration this large would instantly kill any human being.) For a beetle whose mass is 40.00 mg, calculate the magnitude of the force exerted by the beetle on the ground at the beginning of the jump with gravity taken into account. Calculate the magnitude of the force with gravity neglected. Use 9.807 m/s as the value for free-fall acceleration. Problem C 35

AME DATE CLASS 4. In 1994, a Bulgarian athlete named Minchev lifted a mass of 157.5. By comparison, his own mass was only 54.0. Calculate the force acting on each of his feet at the moment he was lifting the mass with an upward acceleration of 1.00 m/s.assume that the downward force on each foot is the same. 5. In 1967, one of the high school football teams in California had a tackle named Bob whose mass was.0. Suppose that after winning a game the happy teammates throw Bob up in the air but fail to catch him. When Bob hits the ground, his average upward acceleration over the course of the collision is 75.0 m/s.(ote that this acceleration has a much greater magnitude than free-fall acceleration.) Find the average force that the ground exerts on Bob during the collision. 6. The whale shark is the largest type of fish in the world. Its mass can be as large as.00 4, which is the equivalent mass of three average adult elephants. Suppose a crane lifts a net with a.00 4 whale shark off the ground. The net is steadily accelerated from rest over an interval of.5 s until the net reaches a speed of 1.0 m/s. Calculate the magnitude of the tension in the cable pulling the net. 7. The largest toad ever caught had a mass of.65. Suppose a toad with this mass is placed on a metal plate that is attached to two cables, as shown in the figure below. If the plate is pulled upward so that it has a net acceleration of.55 m/s,what is magnitude of the tension in the cables? (The plate s weight can be disregarded.) F q 1 = 45 q T,1 = 45 F T, F g = 6 8. In 1991, a lobster with a mass of 0.0 was caught off the coast of ova Scotia, Canada. Imagine this lobster involved in a friendly tug of war with several smaller lobsters on a horizontal plane at the bottom of the sea. Suppose the smaller lobsters are able to drag the large lobster, so that after the large lobster has been moved 1.55 m its speed is 0.550 m/s. If the lobster is initially at rest, what is the magnitude of the net force applied to it by the smaller lobsters? Assume that friction and resistance due to moving through water are negligible. 9. A 0.5 mm wire made of carbon and manganese can just barely support the weight of a 70.0 person. Suppose this wire is used to lift a 45.0 load. What maximum upward acceleration can be achieved without breaking the wire? 36 Holt Physics Problem Workbook

AME DATE CLASS 10. The largest hydraulic turbines in the world have shafts with individual masses that equal 3.18 5. Suppose such a shaft is delivered to the assembly line on a trailer that is pulled with a horizontal force of 81.0 k. If the force of friction opposing the motion is 6.0 k, what is the magnitude of the trailer s net acceleration? (Disregard the mass of the trailer.) 11. An average newborn blue whale has a mass of 3.00 3. Suppose the whale becomes stranded on the shore and a team of rescuers tries to pull it back to sea. The rescuers attach a cable to the whale and pull it at an angle of 0.0 above the horizontal with a force of 4.00 k. There is, however, a horizontal force opposing the motion that is 1.0 percent of the whale s weight. Calculate the magnitude of the whale s net acceleration during the rescue pull. 1. One end of the cable of an elevator is attached to the elevator car, and the other end of the cable is attached to a counterweight. The counterweight consists of heavy metal blocks with a total mass almost the same as the car s. By using the counterweight, the motor used to lift and lower the car needs to exert a force that is only about equal to the total weight of the passengers in the car. Suppose the car with passengers has a mass of 1.600 3 and the counterweight has a mass of 1.00 10 3. Calculate the magnitude of the car s net acceleration as it falls from rest at the top of the shaft to the ground 5.0 m below. Calculate the car s final speed. 13. The largest squash ever grown had a mass of 409. Suppose you want to push a squash with this mass up a smooth ramp that is 6.00 m long and that makes a 30.0 angle with the horizontal. If you push the squash with a force of 080 up the incline, what is a. the net force exerted on the squash? b. the net acceleration of the squash? c. the time required for the squash to reach the top of the ramp? 14. A very thin boron rod with a cross-section of 0.10 mm 0.10 mm can sustain a force of 57. Assume the rod is used to pull a block along a smooth horizontal surface. a. If the maximum force accelerates the block by 0.5 m/s, find the mass of the block. b. If a second force of 4 is applied in the direction opposite the 57 force, what would be the magnitude of the block s new acceleration? Problem C 37

AME DATE CLASS 15. A hot-air balloon with a total mass of.55 3 is being pulled down by a crew tugging on a rope. The tension in the rope is 7.56 10 3 at an angle of 7.3 below the horizontal. This force is aided in the vertical direction by the balloon s weight and is opposed by a buoyant force of 3.10 4 that lifts the balloon upward. A wind blowing from behind the crew exerts a horizontal force of 90 on the balloon. a. What is the magnitude and direction of the net force? b. Calculate the magnitude of the balloon s net acceleration. c. Suppose the balloon is 45.0 m above the ground when the crew begins pulling it down. How far will the balloon travel horizontally by the time it reaches the ground if the balloon is initially at rest? 38 Holt Physics Problem Workbook

Givens 5. m = 155 F T,1 = F T, q 1 = 90 q Solutions F x,net = F T,1 (cos q 1 ) F T, (cos q ) = 0 F y,net = F T,1 (sin q 1 ) + F T, (sin q ) mg = 0 F T,1 [(cos q 1 ) 1 (cos q )] = 0 (cos q 1 ) = cos q = cos(90 q 1 ) = sin q 1 = tan q 1 q 1 = tan 1 () = 63 q = 90 63 =7 F T,1 (sin q 1 ) + F T,1 (sin q ) = mg F T,1 = mg (sin θ 1 ) + 1 (sin θ ) (155 )(9.81 m/s ) F T,1 = = (155 )( 9. 81 m/s ) = (155 )(9.81 m) (sin 63 ) + (sin 7 ) 0.89 + 0. 3 1.1 F T,1 = 1.36 1.36 3 F T, = 6.80 II Additional Practice C 1. v i = 173 km/h v f = 0 km/h x = 0.660 m m = 70.0 a = vf v i [(0 km/h) (173 km/h) ](10 3 m/km) (1 h/3600 s) = x ()(0.660 m) a = 1.75 3 m/s F = ma = (70.0 )( 1.75 3 m/s ) = 1. 10 5 F g = mg = (70.0 )(9.81 m/s ) = 6.87 The force of deceleration is nearly 178 times as large as David Purley s weight.. m =.3 6 a net = 0 m/s 3. m = 40.00 mg = 4.00 5 g = 9.807 m/s a net = (400.0)g a. F net = ma net = F up mg F up = ma net + mg = m(a net + g) = (.3 6 )(0 m/s + 9.81 m/s ) F up =.19 7 =mg b. F down = mg(sin q) a net = F net = F up F down = mg m g(sin q) m m m a net = g(1 sin q) = (9.81 m/s )[1.00 (sin 30.0 )] = 9.81 m/s = 4.90 m/s a net = 4.90 m/s up the incline F net = F beetle F g = ma net = m(400.0) g F beetle = F net + F g = m(400.0 + 1)g = m(401)g F beetle = (4.000 5 )(9.807 m/s )(401) = 1.573 1 F net = F beetle F g = m(400.0) g = (4.000 5 )(9.807 m/s )(400.0) F net = 1.569 1 The effect of gravity is negligible. Section Two Problem Workbook Solutions II Ch. 4 3

II Givens 4. m a = 54.0 m w = 157.5 a net = 1.00 m/s Solutions The net forces on the lifted weight is F w,net = m w a net = F m w g where F is the force exerted by the athlete on the weight. The net force on the athlete is F a,net = F n,1 + F n, F m a g = 0 where F n,1 and F n, are the normal forces exerted by the ground on each of the athlete s feet, and F is the force exerted by the lifted weight on the athlete. The normal force on each foot is the same, so F n,1 = F n, = F n and F =F n m a g Using the expression for F in the equation for F w,net yields the following: m w a net = (F n m a g) m a g F n = m w (a net + g) + m a g F n = m w(a net + g) + m a g (157.5 )(1.00 m/s + 9.81 m/s ) + (54.0 ) = F n = (157.5 )(10.81 m/s ) + (54.0 )(9.81 m/s ) 170 + 5.30 F n = = 3 = 1116 F n,1 F n, = F n = 1116 upward 5. m =.0 a net = 75.0 m/s F net = ma net = F avg mg F avg = m(a net + g) = (.0 )(75.0 m/s + 9.81 m/s ) F avg = (.0 )(84.8 m/s ) = 1.87 4 F avg = 1.87 4 upward 6. m =.00 4 t =.5 v f = 1.0 m/s 7. m =.65 q 1 = q = 45.0 a net =.55 m/s a net = vf v i = (1.0 m/ s 0.0 m/s) = 0.40 m/s t. 5 s F net = ma net = F T mg F T = ma net + mg = m(a net + g) F T = (.00 4 )(0.40 m/s + 9.81 m/s ) F T = (.00 4 )(10.1 m/s ) =.04 5 F T =.04 5 F x,net = F T,1 (cos q 1 ) F T, (cos q ) = 0 F T,1 (cos 45.0 ) = F T, (cos 45.0 ) F T,1 = F T, F y,net = ma net = F T,1 (sin q 1 ) + F T, (sin q ) mg F T = F T,1 = F T, q = q 1 = q F T (sin q) + F T (sin q) = m(a net + g) F T (sin q) = m(a net + g) II Ch. 4 4 Holt Physics Solution Manual

Givens Solutions F T = m (anet + g) = (sin q) (.65 )(.55 m/s = 9.81 m/s ) ()(sin 45.0 ) (.65 )(1.36 m/s ) F T = =3. ()(sin 45.0 ) F T,1 = 3. F T, = 3. 8. m = 0.0 x = 1.55 m v f = 0.550 m/s a net = vf v i (0.550 m/s) (0.00 m/s) = = 9.76 m/s x ()(1.55 m) F net = ma net = (0.0 )(9.76 m/s ) = 1.95 9. m max = 70.0 m = 45.0 F max = m max g = F T F max = (70.0 )(9.81 m/s ) = 687 F net = ma net = F T mg = F max mg a net = F max 687 g = 9.81 m/s = 15.3 m/s 9.81 m/s = 5.5 m/s m 4 5. 0 a net = 5.5 m/s upward II 10. m = 3.18 5 F applied = 81.0 3 F friction = 6.0 3 F net = F applied F friction = (81.0 3 6.0 3 ) F net = 19.0 3 a net = F 3 net 19. 0 = m 3 5. 18 = 5.97 m/s 11. m = 3.00 3 F applied = 4.00 3 q = 0.0 F opposing = (0.10) mg 1. m c = 1.600 3 m w = 1.00 3 y = 5.0 m F net = ma net = F applied (cos q) F opposing a net = a net = F applied(cos q) (0.10) mg m (4.00 3 )(cos 0.0 ) (0.10)(3.00 3 )(9.81 m/s ) 3.00 3 3.76 3 3.53 3.3 a net = = 3 3 3.00 3.00 a net = 7.7 m/s For the counterweight: The tension in the cable is F T. F net = F T m w g = m w a net For the car: F net = m c g F T = m c a net Adding the two equations yields the following: m c g m w g = (m w + m c )a net a net = (m c mw)g = mc + mw (1.600 3 1.00 3 )(9.81 m/s ) 1.600 3 + 1.00 3 (4.00 )(9.81 m/s ) a net = = 1.40 m/s.800 3 Section Two Problem Workbook Solutions II Ch. 4 5

Givens Solutions v f = a ne t y + v i = ( )( 1. 40 m /s ( 5. ) 0 m ) + ( 0 m /s ) v f = 8.37 m/s 13. m = 409 d = 6.00 m q = 30.0 F applied = 080 a. F net = F applied mg(sin q) = 080 (409 )(9.81 m/s )(sin 30.0 ) F net = 080 010 = 70 F net = 70 at 30.0 above the horizontal b. a net = F net 70 = = 0. m/s m 4 09 a net = 0. m/s at 30.0 above the horizontal c. d = v i t + 1 a net t = (0 m/s) t + 1 (0. m/s ) t II 14. a max = 0.5 m/s F max = 57 F app = 4 t = ( ( 0 ). ( 6 0 0 m ) ṁ /s ) = 8 s a. m = F max 57 = a m ax 0. 5 m/s =.3 b. F net = F max F app = 57 4 = 33 a net = F net 33 = m.3 = 0.14 m/s 15. m =.55 3 F T = 7.56 3 q T = 7.3 F buoyant = 3.10 4 F wind = 90 y = 45.0 m a. F x,net =ΣF x = m ax,net = F T (cos q T ) + F wind F x,net = (7.56 3 )[cos( 7.3 )] 90 =.30 3 90 = 1.38 3 F y,net =ΣF y = ma y,net = F T (sin q T ) + F buoyant + F g = F T (sin q T ) + F buoyant mg F y,net = (7.56 3 )[sin( 7.3 )] = 3.10 4 (.55 3 )(9.81 m/s ) F y,net = 7.0 3 + 3.10 4.50 4 = 1. 3 F net = (F x, net ) + ( F y,n et ) = (1.3 8 1 0 3 ) + ( 1. 1 0 3 ) F net = 1. 90 1 0 6 + 1.4 1 0 6 F net = 3. 3 1 0 6 = 1.8 3 q = tan 1 Fy, net Fx, net = tan 1 3 1. 1 1. 38 0 3 10 q = 41 F net = 1.8 3 at 41 below the horizontal b. a net = F 3 net 1.8 = m 3.55 a net = 0.71 m/s c. Because v i = 0 y = 1 a y,net t x = 1 a x,net t x = a x, net a y = net (cos q) y y = ay, net anet (sin q) ta n q 45.0 m x = = 5 m ta n( 41 ) II Ch. 4 6 Holt Physics Solution Manual