Unformatted text preview: CHAPTER 5 FORCE AND MOTION
ActivPhysics can help with these problems: All activities in Section 2, Forces and Motion Section 54: Newton's Second Law Problem
1. A subway train has a mass of 1.5x106 kg. What force is required to accelerate the train at 2.5 m/s2? Solution
Assume that the seat belt holds the passenger firmly to the seat, so that the passenger also stops in 0.14 s without incurring any secondary impact. Then the passenger's average acceleration is aav = (0  vo)/t, and the average net force on the passenger, while coming to rest, is Fav = maav = mvo/t = (60kg)(1l0/3.6)(m/s)/(0.14 s) = 13.1 kN, or about 1.5 tons. (Here, we used the onedimensional form of Newton's second law. The minus sign indicates that the force is opposite to the direction of the initial velocity. It is reasonable to assume that this is a component of the force exerted by the seatbelt.) Solution
F = ma = (1.5x106 kg)(2.5 m/s2) = 3.75 MN. (This is the magnitude of the net force acting; see Table 11 for SI prefixes.) Problem
2. A railroad locomotive with a mass of 6.1x 104 kg can exert a force of 1.2xl05 N. At what rate can it accelerate (a) by itself and (b) when pulling a L4x106kg train? Problem
5. In an xray tube, electrons are accelerated to speeds on the order of 108 mis, then slammed into a target where the); come to a stop in about 1O18S. Estimate the average stopping force on each electron. Solution
Ignoring the probable presence of other forces, we can apply Equation 53 to find (a) a = F/m = (1.2x105 N)/(6.1x104 kg) = 1.97 m/s2, and (b) a = (1.2x105N)/(1.46x106 kg) = 8.21 cm/s2• I Solution
The magnitude of the average force is P = mii = m l6.v/6.tl = (9.11 X 1031 kg)(108 m/s)/(1O18 s) ~ 9xlO5 N. Compared to the TV tube in Example 52, the electron in an xray tube experiences a force billions of times greater. It is a. result of the violence of this interaction that x rays, called bremsstrahlung, are emitted (see Problem 245). Problem
3. A small plane starts down the runway with acceleration 7.2 m/s2• If the force provided by its engine is 1.1x 104 N, what is the plane's mass? II
t Solution
If we assume that the runway is horizontal (so that the vertical force of gravity and the normal force of the surface cancel) and neglect aerodynamic forces (which are small just after the plane begins to move) then the net force equals the engine's thrust and is parallel to the acceleration. The horizontal component of Equation 53 gives the airplane's mass, m= F/a = (1.1xl04 N)/(7.2 m/s2) = 1.53x 103 kg. Problem
6. A 280kg crate is secured in a truck with ropes running horizontally forward and backward, as shown in Fig. 533. What is the maximum tension force these ropes must be able to withstand if the most rapid deceleration anticipated is 6.5 m/s2? Assume the tension is initially zero. I
I
f t ~ Problem
4. A car leaves the road traveling at 110 km/h and hits a tree, coming to a complete stop in 0.14 s. What average force does a seatbelt exert on a 60kg passenger during this collision? FIGURE 5":33Problem 6. CHAPTER 5 71 Solution Suppose that the motion of the truck is also horizontal, positive to the right, in the direction shown in Fig. 533, and consider only the horizontal forces. The maximum anticipated net horizontal force on the crate is Fnet == mamax = (280 kg)(6.5 m/s2) = 1.82 kN, where the minus sign indicates a deceleration, or force to the left. The ropes can exert only tensile forces; therefore, the lefthand rope can only pull to the left (Ti < 0), and the righthand rope can only pull to the right (Tr > 0). If there is no friction between the crate and the truck bed, the ropes must supply all the horizontal force on the crate, so Fnet = Ti + Tr = Tr ITtI; that is, the maximum difference in tension between the ropes is 1.82 kN. (We wrote this equation in terms of the absolute values.because that is what is normally meant by the tension in a rope.) If the ropes are initially under zero tension, then all the tension when decelerating is in the lefthand rope, and tension in the righthand rope is zero. maav = m /lv/ /It = (0.170 kg)(50 m/s  0) + (2.5 ms)= 3.40 kN. Problem
9. By how much does the force required to stop a car increase if the initial speed is doubled and the stopping distance remains the same? Solution The average net force on a car of given mass is proportional to the average acceleration, Fav '" aav. To stop a car in a given distance, (x  xo), aav = (0  v5)/2(x  xo), so Fav '" v5. Doubling Vo quadruples the magnitude of Fav, a fact that is important to remember when .driving cat high speeds. Problem
10. A car moving at 70 km/h collides with a concrete bridge support. The bridge support is unaffected, but the front of the car is compressed by 0.94 m. What average force must a seat belt exert in order to restrain a 75kg passenger during this collision? Solution The average acceleration of the car, while stopping in a horizontal distance x  Xo, is aav = (0  vs) + 2(x  xo) = (70 m/3.6 s)2/2(0.94 m) = 201 m/s2, or about 21g. A properly functioning seat belt will constrain a passenger to the seat and prevent any secondary collision with the interior compartment of the automobile. Then the passenger's acceleration is also 201 m/s2, and the magnitude of the average horizontal net force on her/him is simply Fnet = maav = (75 kg)(201 m/s2) = 15.1 kN. This force is supplied mainly by the seat belt. Problem
7. Object A accelerates at 8.1 m/s2 when a 3.3N force is applied. Object B accelerates at 2.7 m/i when the same force is applied. (a) How do the masses of the two objects compare? (b) If A and B were stuck together and accelerated by the 3.3N force, what would be the acceleration of the composite object? Solution In this idealized onedimensional situation, the applied force of F = 3.3 N is the only force acting. (a) When applied to either object, Newton's secorid law gives F = mAaA and F = mBaB, so mB/mA = aA/aB = (8.1 m/s2)/(2.7 m/s2) = 3. (For constant net force, mass is inversely proportional to acceleration.) (b) When F is applied to the combined object, F = (mA + mB)a. Since F = mAaA, and mB = 3mA, one finds a = F/(mA + mB) = mAaA/4mA = ~(8.1 m/s2) = 2.03 m/s2. (Note: It was not necessary to calculate the masses, which are mA = (3.3 N) + (8.1 m/s2) = 0.407 kg, and mB = (3.3 N) + (2.7 m/s2) = 1.22 kg.) Problem
11. The maximum braking force of a 1400kg car is about 8.0 kN. Estimate the stopping distance when the car is traveling (a) 40 lan/hj (b) 60 km/hj (c) 80 km/hj (d) 55 mi/h. Solution The maximum braking acceleration is a = F/m = 8.0x103 N/1400 kg = 5.71 m/s2. (We expressed the braking force as negative because it is opposite to the direction of motion.) (a) On a straight horizontal road, acai traveling at a velocity of VOx = 40 km/h can stop in a minimum distance found from Equation 211 and the maximum deceleration just calculated: x  xo = v5x/2a = (40 m/3.6 s)2+ 2('5.71 m/i) = 10.8 m. For the other initial Problem
8. A hockey player strikes a 170g puck, accelerating it from rest to 50 m/s. If the hockey stick is in contact with the puck for 2.5 ms, what is the average force applied by the stick? Solution The stick is responsible for essentially all the average net force on the puck during impact, so Fav = 72 CHAPTER 5 velocities, the stopping distance is (b) 24.3 m, (c) 43.2 rri, and (d) 52.9 m. Problem
12. As a function of time, the velocity of an object of mass m is given by v = bt2i + (d + d)j, where b, c, and d are constants with appropriate units. What is the force acting on the object, as a function of time? average force of air on the parachute? Assume the parachute provides essentially all the stopping force. Solution
To stop a 380okg object traveling at 240 km/h in 170 m requires an average force of F = rna = ml v6l2(x  xo)/ = (3800 kg)(240 m/3.6 S)2..;2(170 m) = 49.7 kN. Solution
From the discussion leading to Equation 53, F net = mdv/dt = m(2bti+cj). Problem
15. A 1.25kg object is moving in the x direction at 17.4 m/s. 3.41 s later, it is moving at 26.8 m/s at 34.0° to the :vaxis. What are the magnitude and direction of the force applied during this time? Problem
13. A car moving at 50 km/hcollides with a truck, and the front of the car is crushed 1.1 m as it comes to a complete stop. The driver is wearing a seatbelt, but the passenger is not. The passenger, obeying Newton's first law, keeps moving and slams into the dashboard after the car has stopped. If the dashboard compressed 5.0 cm on impact, find and compare the forces exerted on the driver by the seat belt and on the passenger by the dashboard. Assume the two have the same 65kg mass. Solution
Newton's second law says that the average force acting is equal to the rate of change of momentum, F av m~v/~t (as explained following Equation 52). The initial velocity is 17.4i mis, and the final velocity is (26.8 m/s)(i cos 34° + j sin 34°) = (22.2i + 15.0j) mis, so Fav = (1.25 kg)[(22.2 17.4) i + 15.0j](m/s)/(3.41 s) = (1.77i+5.49j) N. This has magnitude 5.77 N and direction 72.2° CCW,to the xaxis. = ~. Solution
The seatbelt constrains the driver to have the same average acceleration as the car, while stopping. Assuming the passenger compartment stays intact and is stopped after moving a horizontal distance of ~Xd = 1.1 m, we can estimate the driver's average acceleration from Equation 211 as ad = v6l2~Xd. Then the magnitude of the average force stopping the driver is Fd = mladl =~(65 kg)(50 m/3.6 S)2..;(1.1 m) = 5.70 kN(over half a ton!), which is exerted mostly by the seat belt. A passenger striking a stationary dashboard with the same speed Vo, which yields a horizontal distance ~xp = 0.05 m before bringing her/him to a stop (probably permanently), experiences an average acceleration of ap = v5 ..;2~xp = (~Xd/~xp)ad = (Ll m/0.05 m)ad = 22.0ad. Therefore, the average force stopping the passenger is 22 times greater than that for the driver, or 22x5.70 kN = 125 kN. Buckleup for safetyit's a law of physics! Section 55: Mass and Weight: The Force of Gravity Problem
16. Show that the units of acceleration can be written N/kg. Why would it make sense to state 9 as 9.8 N/kg when talking about mass and weight? Solution
From Newton's second law, [acceleration] = [Forcel/[massj = N/kg ([... ] means "units of ... "). 9 = 9.8 N/kg is the conversion factor from mass to the everyday meaning of weight (Le., the force of a scale reading at reston the surface of the Earth). Problem
17. My spaceship crashes on one of the Sun's nine planets. Fortunately, the ship's scales are intact, and show that my weight is 532 N. If 1 know my mass to be 60 kg, where am I? Hint: Consult AppeIiaix E. Problem
14. A 3800kgjet touches down at 240 km/h on the deck of an aircraft carrier, and immediately deploys a parachute to slow itself down. If the plane comes to a stop in 170 m, what is the Solution
The surface gravity of the planet is 9 = W/m = 532 N/60 kg = 8.87 m/s2, precisely the value for Venus in Appendix E. CHAPTER 5 73 Problem
18. If I can barely lift a 35kg concrete block on Earth, how massive a block can I lift on the moon? value. What is the weight of a 68~kgastronaut in a shuttle at this altitude? Solution
The magnitude of the Earth's gravitational force on the astronaut at this altitude is Fgrav = mg(r) = (68 kg)(0.93x9.8 rn/s2) = 620N, where g(r) is the gravitational acceleration at the appropriate distance, r, from the center of the Earth. Of course, if the astronaut is orbiting with the shuttle, Le., in free fall around the Earth, his or her "weight" is zero. (An operational definition of ''weight'' is the force read on a scale at rest relative to the object being weighed.) Solution
To lift a 35kg block on Earth requires a muscular force of at least its weight, W = mg = (35 kg) x (9.8 m/s2) = 343 N. The same force on the moon could lift a mass m = W/gmoon = 343 N/(1.62 m/s2) = 212 kg. Problem
19. A cereal box says "net weight 340 grams." What is the actual weight (a) in SI units? (b) in ounces? Solution
(a) The actual weight (equal to the force of gravity at rest on the surface of the Earth) is mg = (0.340 kg)x 2 (9.81 m/s ) = 3.33 N. (b) With reference to Appendix C, (3.33 N)(0.2248x 16 oz/N) = 12.0 oz. (The word "net" in net weight means just the weight of the contents; gross weight includes the weight of the container, etc. This may be compared with the use of the word in net force, which means the sum of all the forces or the resultant force. A net weight, profit, or amount is the resultant after all corrections. have been taken into account.) Problem
23. A neutron star is a fantastically dense object with the mass of a star crushed into a region about 10 km in diameter. If my mass is 75 kg, and if I would weigh 5.8x1014 N on a certain neutron star, what is the acceleration of gravity 011 the neutron star? Solution
If we define weight on a neutron star analogously to its definition on Earth, the surface gravity of the neutron star is an enormous g = W/m = (5.8x 1014 N) + (75 kg) = 7.73x1012 m/s2, nearly 1012 times g. Problem
20. I weigh 735 N. What's my mass? Section 56: Problem Adding Forces Solution
m = Wig = 735 N/(9.81 m/s ) = 74.9 kg (for an object at rest on the surface of the Earth, where 2 9 = 9.81 m/s ). .
2 24. A 50kg parachute jumper descends at a steady 40 km/h. What is the force of air on the parachute? Solution
A steady descent means that the vertical acceleration is zero. From Newton's second law, the vertical net force is also zero. The force of gravity (equal to the weight of the parachutist) and the force of the air are the only vertical forces acting, so (with positive upward) Fair  W = 0, or Fair = (50 kg)(9.8 m/s2) = 490 N.
\ Problem
21. A bridge specifies a maximum load of 10 tons. What's the maXimum mass, in kilograms, that the bridge can carry? Solution
The conversion between mass and (ordinary) weight' is m = Wig. Because the English unit of mass (the slug) is rarely used, the direct equivalence between mass in SI units and weight (force) in English units is usually given, as in Appendix C. Thus 10 tons = 2x104 lb is equivalent to the weight of (2x104 Ib)(0.4536 kg/lb) = 9.07x103 kg. Problem
25. A rope can withstand a maximum tension force of 450 N before breaking. The rope is used to pull a 32kg bucket of water upward. What is the maximum upward acceleration if the rope is not to break? Problem
22. The gravitational acceleration at a typical space shuttle altitude is about 93 percent of its surface Solution
Assume that the only vertical forces acting on the bucket are the upward (positive) force exerted by the 74 CHAPTER 5 rope, .Prope :5 450 N, and the downward force of gravity, Fgrav = mg. (Air resistance is ignored.) The vertical component of Equation 53 applied to the bucket is Fnet = Frape + Fgrav = ma, so a = 2 (Frope  mg)/m :5 (450 N/32 kg)  9.8 m/s = 2 4.26 m/s , which is the maximum upward acceleration aA::hievable with this rope. (Note that if the tensile strength of the rope were less than the weight of the bucket, 314 N, the maximum acceleration would have been downward; see Problem 60.) or Flift = m(g + ay).(a) At constant altitude, ay = 0, and Flift = mg = (4.5 X 1Q5kg)(9.8 m/i) = 4.41 MN, the airplane's weight. (b) If ay = 1.1 m/i, then Flift = (4.41 MN)(l + 1.1/9.8) = 4.91 MN. Problem
29. An airplane encounters sudden turbulence, and you feel momentarily lighter. If your apparent weight seems to be about 70% of your normal weight, what are the magnitude and direction of the plane's acceleration? Problem
26. A 930kg motorboat aA::celerates away from a dock 2 at2.3 m/s • Its propeller provides a thrust force of 3.9 kN. What is the drag force exerted by the water on the boat? Solution
The vertical forces acting on you are gravity downward (mg) and the normal force of your seat (N 2: 0). The latter is what you experience as your apparent weight. During the turbulence, the vertical component of Newton's second law gives Fnet y = mg + N = mg + 70%mg = may; therefore ' 2 ay = 30%g = 2.94 m/s . If the horizontal components of the acceleration are zero, then the plane's aA::celeration 2.94 m/s2 downward. is Solution
The horizontal forces acting on the boat are the thrust and the oppositely directed drag force (take the positive :vaxis in the direction of the aA::celeration). Then the horiwntal component of Newton's second law gives Fnet,z = Fthrust,z + Fdrag,z = 3.9 kN + 2 Fdrag,z = maz = (930 kg)(2.3 m/s ), from which one finds Fdrag,z = 2.14 kN  3.9 kN = 1.76 kN. (As expected, the horizontal component of the drag force is negative, Le., opposite to the direction of the aA::celeration. ) Problem
30. A 74kg tree surgeon rides a "cherry picker" lift (Fig. 534) ,to reach the upper branches of a tree. What force does the bucket of the lift exert on the surgeon when the bucket is (a) at rest; (b) moving upward at a steady 2.4 m/s; (c) moving downward at a steady 2.4 m/s; (d) accelerating upward at 2 1.7 m/s ; (e) aA::celerating downward at 1.7 m/s2? Problem
27. An elevator aA::celerates ownward at 2.4 m/s2• d .What force doe; the floor of the elevator exert on a 52kg passenger? Solution
We assume that the only vertical forces acting on the tree surgeon are those given, namely, the force of gravity, mg acting downward, and the normal force of the bucket, N 2: 0 aA::ting upward. Then Fnet,y = mg + N = may, or N = m(g + ay). (a), (b) and (c) When the acceleration of the tree surgeon. is zero, the normal force is equal (in magnitude) to the weight, N = mg = (74 kg)(9.8 m/s2) = 725 N. (d) If ay = 2 +1.7 tn/s , N = (74 kg)(9.8 + 1.7) m/s2 = 851 N, while if (e) ay = 1.7 m/i,N = 599 N. Solution
The passenger also accelerates downward with 2 ay = 2.4 m/s (yaxis positive upward), so the vertical component of tJIe net force on the passenger is Fnet,y = may. The only vertical forces aA::ting n the o passenger are the force of gravity, Fg,y =  W = mg, and the normal force of the floor, Fnorm,y = N. Therefore, Fnet y = mg + N = may, or , 2 N = m(g + ay) = (52 kg)(9.8  2.4) m/s = 385 N. Problem
28. What is the vertical lifting force on a 747 jetliner when the plane is (a) flying at constant altitude and (b) aA::celerating pward at 1.1 m/s2? The u aircraft's mass is 4.5x105 kg. Problem
31. At liftoff, a space shuttle with 2.0x 106 kg total masS undergoes an upward aA::celeration f 0.60g. o (a) What is the total thrust force developed by its engines? (b) What force does the seat exert on a 60kg astronaut during liftoff? Solution
The vertical forces on the airplane are lift and gravity, so with the yaxis upward, Fnet,y = Flift mg= may, CHAPTER 5 75 Solution
(a) At liftoff, the only significant vertical forces on the space shuttle are gravity (Fg,y = mg downward) and the thrust (Fy > 0 upward). (Air resistance can be neglected because the initial velocity is zero.) Therefore, Fnet,y = Fy  mg = may, or Fy = 2 meg + ay) = mg(l + 0.6) = (2x106 kg)(9.8 m/s ) X (1.6) = 31.4 MN. (b) The vertical forces on an astronaut are her Ihis weight and the normal force of the seat back (N > 0 upward; the astronauts are seated facing upward at liftoff). The acceleration of the astronaut is the same as that of the shuttle, so with reasoning analogous to part (a) above, N = meg + ay) = (60 kg)(9.8 m/s2)(1.6) = 941 ~. Problem
34. A 2.50kg object is moving along the :z;..axis t a 1.60 m/s. As it passes the origin, two forces F1 and F2 are applied, both in the y direction (+ or ). Fl=15j N. The forces are applied for 3.00 s, after which the object is at the point x = 4.80 m, y = 10.8 m. Find F2. Solution
The motion is in the xy plane, and the object's acceleration a= (Fl +F2)/m has only a 'y component, ax = 0, ay = (Fly + F2y)/m. For constant acceleration, the given initial conditions (xo = Yo 0, Vox = 1.60 mis, vOy = 0) imply y(3 s) = 10.8 m = !ay(3 s)2 or ay = 2.4 m/s2. Then F2y , 2 may  Fly = (2.5 ...
View
Full Document
 Spring '08
 WORMER
 Physics, Force, Mass, kg

Click to edit the document details