Solution:<\/strong><\/span>
\n<\/p>\nChapter 6 Applications Of Newton’s Laws Q.1P<\/strong>
\n
\nSolution:<\/strong><\/span>
\n<\/p>\nChapter 6 Applications Of Newton’s Laws Q.2CQ<\/strong>
\nIn the Jurassic Park sequel, The Lost World, a man tries to keep a large vehicle from going over a cliff by connecting a cable from his Jeep to the vehicle. The man then puts the Jeep in gear and spins the rear wheels. Do you expect that spinning the tires will increase the force exerted by the Jeep on the vehicle? Why or why not?
\nSolution:<\/strong><\/span>
\nNo
\nThe man puts the jeep in gear and spins the rear wheels, but spinning will not provide the friction needed to rise above. Spinning the wheels actually decrease the force exerted by the jeep because the force exerted by the spinning wheels is kinetic friction, and the coefficient of kinetic friction is generally less than the coefficient of static friction.<\/p>\nChapter 6 Applications Of Newton’s Laws Q.2P<\/strong>
\nPredict\/Explain Two drivers traveling side-by-side at the same speed suddenly see a deer in the road ahead of them and begin braking. Driver 1 stops by locking up his brakes and screeching to a halt; driver 2 stops by applying her brakes just to the verge of locking, so that the wheels continue to turn until her car comes to a complete stop. (a) All other factors being equal, is the stopping distance of driver 1 greater than, less than, or equal to the stopping distance of driver 2? (b) Choose the best explanation from among the following:
\nI. Locking up the brakes gives the greatest possible braking force.
\nII. The same tires on the same road result in the same force of friction.
\nIII. Locked-up brakes lead to sliding (kinetic) friction, which is less than rolling (static) friction.
\nSolution:<\/strong><\/span>
\n(a) The stopping distance of driver 1 is greater than the stopping distance of driver 2.
\n(b) For driver 2 the force stopping the car is the static friction force. And for driver 1 the force stopping the car is the kinetic friction force. But the static friction force is greater than the kinetic friction force. Therefore the driver 1 travels greater than the driver 2. So option III is the best explanation.<\/p>\nChapter 6 Applications Of Newton’s Laws Q.3CQ<\/strong>
\nWhen a traffic accident is investigated, it is common for the length of the skid marks to be measured. How could this information be used to estimate the initial speed of the vehicle that left the skid marks?
\nSolution:<\/strong><\/span>
\nBraking distance depends on the initial speed and the kinetic friction. If the braking distance and the kinetic friction are known, then the initial speeds of the car can be found.<\/p>\nChapter 6 Applications Of Newton’s Laws Q.3P<\/strong>
\nAbaseball player slides into third base with an initial speed of 4.0 m\/s. If the coefficient of kinetic friction between the player and the ground is 0.46, how far does the player slide before coming to rest?
\nSolution:<\/strong><\/span>
\n
\n<\/p>\nChapter 6 Applications Of Newton’s Laws Q.4CQ<\/strong>
\nIn a car with rear-wheel drive, the maximum acceleration is often less than the maximum deceleration. Why?
\nSolution:<\/strong><\/span>
\nThe maximum acceleration is determined by the normal force exerted on the drive wheels. If the engine of the car is in the front and the drive wheels are in the rear, the normal force is less than it would be with front-wheel drive. During braking, however, all four wheels participate \u2013 including the wheels that sit under the engine.<\/p>\nChapter 6 Applications Of Newton’s Laws Q.4P<\/strong>
\nA child goes down a playground slide with an acceleration of 1.26 m\/s2. Find the coefficient of kinetic friction between the child and the slide if the slide is inclined at an angle of 33.0\u00b0 below the horizontal.
\nSolution:<\/strong><\/span>
\n
\n
\n<\/p>\nChapter 6 Applications Of Newton’s Laws Q.5CQ<\/strong>
\nA train typically requires a much greater distance to come to rest, for a given initial speed, than does a car. Why?
\nSolution:<\/strong><\/span>
\nThe frictional force is responsible for moving the object when the brakes or the driving force are applied. In the case of the train, the frictional force between the rails and the wheels are comparatively low because both are smooth surfaces.
\nIn the case of the car, the road and the rubber tires are both irregular surfaces, so the frictional force is comparatively greater. Therefore, the car stops at a shorter distance than the train.<\/p>\nChapter 6 Applications Of Newton’s Laws Q.5P<\/strong>
\nHopping into your Porsche, you floor it and accelerate at 12 m\/s2 without spinning the tires. Determine the minimum coefficient of static friction between the tires and the road needed to make this possible.
\nSolution:<\/strong><\/span>
\n<\/p>\nChapter 6 Applications Of Newton’s Laws Q.6CQ<\/strong>
\nGive some everyday examples of situations in which friction is beneficial.
\nSolution:<\/strong><\/span>
\nFrictional force is beneficial in the following cases.
\n(1) Without friction, we cannot walk
\n(2) Without friction, a car cannot run on the road
\n(3) Without friction, we cannot hammer the nails inside the walls<\/p>\nChapter 6 Applications Of Newton’s Laws Q.6P<\/strong>
\nWhen you push a 1.80-kg book resting on a tabletop, it takes 2.25 N to start the book sliding. Once it is sliding, however, it takes only 1.50 N to keep the book moving with constant speed. What are the coefficients of static and kinetic friction between the book and the tabletop?
\nSolution:<\/strong><\/span>
\n<\/p>\nChapter 6 Applications Of Newton’s Laws Q.7CQ<\/strong>
\nAt the local farm, you buy a flat of strawberries and place them on the backseat of the car. On the way home, you begin to brake as you approach a stop sign. At first the strawberries stay put, but as you brake a bit harder, they begin to slide off the seat. Explain.
\nSolution:<\/strong><\/span>
\nAs you brake harder, your car has a greater acceleration. The greater the acceleration of the car, the greater the force required to give the flat strawberries the same acceleration. When the required force exceeds the maximum force of static friction, the strawberries begin to slide.<\/p>\nChapter 6 Applications Of Newton’s Laws Q.7P<\/strong>
\nIn Problem, what is the frictional force exerted on the book when you push on it with a force of 0.75 N? When you push a 1.80-kg book resting on a tabletop, it takes 2.25 N to start the book sliding. Once it is sliding, however, it takes only 1.50 N to keep the book moving with constant speed. What are the coefficients of static and kinetic friction between the book and the tabletop?
\nSolution:<\/strong><\/span>
\n
\n<\/p>\nChapter 6 Applications Of Newton’s Laws Q.8CQ<\/strong>
\nIt is possible to spin a bucket of water in a vertical circle and have none of the water spill when the bucket is upside down. How would you explain this to members of your family?
\nSolution:<\/strong><\/span>
\nWhen we rotate the bucket vertically, a centripetal force comes to counter the weight of the water that falls down. A sufficiently high speed of rotation gives a sufficient force opposite to the mouth of the bucket, so the water does not fall at all.<\/p>\nChapter 6 Applications Of Newton’s Laws Q.8P<\/strong>
\n
\nSolution:<\/strong><\/span>
\n
\n<\/p>\nChapter 6 Applications Of Newton’s Laws Q.9CQ<\/strong>
\nWater sprays off a rapidly turning bicycle wheel. Why?
\nSolution:<\/strong><\/span>
\n<\/p>\nChapter 6 Applications Of Newton’s Laws Q.9P<\/strong>
\nA tie of uniform width is laid out on a table, with a fraction of its length hanging over the edge. Initially, the tie is at rest. (a) If the fraction hanging from the table is increased, the tie eventually slides to the ground. Explain. (b) What is the coefficient of static friction between the tie and the table if the tie begins to slide when one-fourth of its length hangs over the edge?
\nSolution:<\/strong><\/span>
\n<\/p>\nChapter 6 Applications Of Newton’s Laws Q.10CQ<\/strong>
\nCan an object be in equilibrium if it is moving? Explain.
\nSolution:<\/strong><\/span>
\nAnswer: Yes
\nExplanation:
\nIf a body moving with constant velocity, it acted upon by zero net force. The body is said to be in equilibrium, if net force acting on it is zero. Therefore, object can be in equilibrium if it is moving with a constant velocity.<\/p>\nChapter 6 Applications Of Newton’s Laws Q.10P<\/strong>
\n
\nSolution:<\/strong><\/span>
\n
\n
\n<\/p>\nChapter 6 Applications Of Newton’s Laws Q.11CQ<\/strong>
\nIn a dramatic circus act, a motorcyclist drives his bike around the inside of a vertical circle. How is this possible, considering that the motorcycle is upside down at the top of the circle?
\nSolution:<\/strong><\/span>
\nThe motorcyclist drives his bike around the inside vertical circle at a very high speed. Because of this high speed, sufficient centripetal force appears away from the center and is much higher than the weight of the motorcyclist and his bike when it is upside down.<\/p>\nChapter 6 Applications Of Newton’s Laws Q.11P<\/strong>
\n
\nSolution:<\/strong><\/span>
\n
\n<\/p>\nChapter 6 Applications Of Newton’s Laws Q.12CQ<\/strong>
\nThe gravitational attraction of the Earth is only slightly less at the altitude of an orbiting spacecraft than it is on the Earth\u2019s surface. Why is it, then, that astronauts feel weightless?
\nSolution:<\/strong><\/span>
\nAnswer:
\nIn this case, two astronauts are in the constant-free fall motion as they are in orbiting.
\nFor constant free-fall motion, the net gravitational force of attraction acting on the astronauts is zero. Hence, astronauts feel weightless. This is just resembles to the case, if elevator drops downward in free fall motion, you will feel weightless inside the elevator.<\/p>\nChapter 6 Applications Of Newton’s Laws Q.12P<\/strong>
\nA 48-kg crate is placed on an inclined ramp. When the angle the ramp makes with the horizontal is increased to 26\u00b0, the crate begins to slide downward. (a) What is the coefficient of static friction between the crate and the ramp? (b) At what angle does the crate begin to slide if its mass is doubled?
\nSolution:<\/strong><\/span>
\n
\n<\/p>\nChapter 6 Applications Of Newton’s Laws Q.13CQ<\/strong>
\nA popular carnival ride has passengers stand with their backs against the inside wall of a cylinder. As the cylinder begins to spin, the passengers feel as if they are being pushed against the wall. Explain.
\nSolution:<\/strong><\/span>
\nDuring the carnival ride, when the cylinder begins to spin, its centripetal force is exerted on the passengers. This force, which is radially inward, is supplied by the wall of the cylinder.<\/p>\nChapter 6 Applications Of Newton’s Laws Q.13P<\/strong>
\nA 97-kg sprinter wishes to accelerate from rest to a speed of 13 m\/s in a distance of 22 m. (a) What coefficient of static friction is required between the sprinter\u2019s shoes and the track? (b) Explain the strategy used to find the answer to part (a).
\nSolution:<\/strong><\/span>
\n
\n<\/p>\nChapter 6 Applications Of Newton’s Laws Q.14CQ<\/strong>
\nReferring to Question, after the cylinder reaches operating speed, the floor is lowered away, leaving the passengers \u201cstuck\u201d to the wall. Explain.
\n(Answers to odd-numbered Conceptual Questions can be found in the back of the book.) A popular carnival ride has passengers stand with their backs against the inside wall of a cylinder. As the cylinder begins to spin, the passengers feel as if they are being pushed against the wall. Explain
\nSolution:<\/strong><\/span>
\nReaching the operating speed, the centripetal force acting on the man is sufficient to counter his weight, which is responsible for the fall. Thus, even without the base, the man sticks to the wall.<\/p>\nChapter 6 Applications Of Newton’s Laws Q.14P<\/strong>
\nCoffee To Go A person places a cup of coffee on the roof of her car while she dashes back into the house for a forgotten item. When she returns to the car, she hops in and takes off with the coffee cup still on the roof. (a) If the coefficient of static friction between the coffee cup and the roof of the car is 0.24, what is the maximum acceleration the car can have without Causing the cup to slide? Ignore the effects of air resistance. (b) What is the smallest amount of time in which the person can accelerate the car from rest to 15 m\/s and still keep the coffee cup on the roof?
\nSolution:<\/strong><\/span>
\n<\/p>\nChapter 6 Applications Of Newton’s Laws Q.15CQ<\/strong>
\nYour car is stuck on an icy side street. Some students on their way to class see your predicament and help out by sitting on the trunk of your car to increase its traction. Why does this help?
\nSolution:<\/strong><\/span>
\nStudents sitting on the trunk of the car increase the normal force between the tires and the road. The force of friction is directly proportional to the normal, so this increases the frictional force enough so that the car moves.<\/p>\nChapter 6 Applications Of Newton’s Laws Q.15P<\/strong>
\nForce Times Distance I At the local hockey rink, a puck with a mass of 0.12 kg is given an initial speed of v = 5.3 m\/s. (a) If the coefficient of kinetic friction between the ice and the puck is 0.11, what distance d does the puck slide before coming to rest? (b) If the mass of the puck is doubled, does the frictional force F exerted on the puck increase, decrease, or stay the same? Explain. (c) Does the stopping distance of the puck increase, decrease, or stay the same when its mass is doubled? Explain. (d) For the situation considered in part (a), show that . (The significance of this result will be discussed in Chapter 7, where we will see that is the kinetic energy of an object.)
\nSolution:<\/strong><\/span>
\n
\n<\/p>\nChapter 6 Applications Of Newton’s Laws Q.16CQ<\/strong>
\nThe parking brake on a car causes the rear wheels to lock up. What would be the likely consequence of applying the parking brake in a car that is in rapid motion? (Note: Do not try this at home.)
\nSolution:<\/strong><\/span>
\nIf the parking brake is applied while the car is in motion, and the rear wheels begin to skid across the pavement. This means that the friction acting on the rear wheels is kinetic friction. This kinetic friction is smaller than the static friction experienced by the front wheels. As a result, the rear wheels overtake the front wheels causing the car to spin around, and the rear wheels begin to move first.<\/p>\nChapter 6 Applications Of Newton’s Laws Q.16P<\/strong>
\nForce Times Time At the local hockey rink, a puck with a mass of 0.12 kg is given an initial speed of v0 = 6.7 m\/s. (a) If the coefficient of kinetic friction between the ice and the puck is 0.13, how much time f does it take for the puck to come to rest? (b) If the mass of the puck is doubled, does the frictional force F exerted on the puck increase, decrease, or stay the same? Explain. (c) Does the stopping time of the puck increase, decrease, or stay the same when its mass is doubled? Explain. (d) For the situation considered in part (a), show that Ft = mv0. (The significance of this result will be discussed in Chapter 9, where we will see that mv is the momentum of an object.)
\nSolution:<\/strong><\/span>
\n
\n
\n<\/p>\nChapter 6 Applications Of Newton’s Laws Q.17CQ<\/strong>
\nThe foot of your average gecko is covered with billions of tiny hair tips\u2014called spatulae\u2014that are made of keratin, the protein found in human hair. A subtle shift of the electron distribution in both the spatulae and the wall to which a gecko clings produces an adhesive force by means of the van der Waals interaction between molecules. Suppose a gecko uses its spatulae to cling to a vertical windowpane. If you were to describe this situation in terms of a coefficient of static friction, \u00b5s, what value would you assign to \u00b5s? Is this a sensible way to model the gecko\u2019s feat? Explain.
\nSolution:<\/strong><\/span>
\nThe normal force exerted on a gecko by the vertical wall is zero. For the gecko to stay in place, its static friction must exert an upward force equal to the gecko\u2019s weight. In order for this to happen, the normal force should be zero, and the coefficient of static friction should be infinite.<\/p>\nChapter 6 Applications Of Newton’s Laws Q.17P<\/strong>
\n
\nSolution:<\/strong><\/span>
\n
\n
\n<\/p>\nChapter 6 Applications Of Newton’s Laws Q.18CQ<\/strong>
\nDiscuss the physics involved in the spin cycle of a washing machine. In particular, how is circular motion related to the removal of water from the clothes?
\nSolution:<\/strong><\/span>
\nAs the basket within a washing machine rotates, the clothes collect on the rim of the basket because of the centripetal force acting on the rotation. The basket exerts an inward force on the clothes, causing them to follow a circular path. The water contained in the clothes, however, is able to pass through the holes of the basket where it can be drained from the machine.<\/p>\nChapter 6 Applications Of Newton’s Laws Q.18P<\/strong>
\nThe coefficient of kinetic friction between the tires of your car and the roadway is \u00b5. (a) If your initial speed is v and you lock your tires during braking, how far do you skid? Give your answer in terms of v, \u00b5, and m, the mass of your car. (b) If you double your speed, what happens to the stopping distance? (c) What is the stopping distance for a truck with twice the mass of your car, assuming the same initial speed and coefficient of kinetic friction?
\nSolution:<\/strong><\/span>
\n
\n<\/p>\nChapter 6 Applications Of Newton’s Laws Q.19CQ<\/strong>
\nThe gas pedal and the brake pedal are capable of causing a car to accelerate. Can the steering wheel also produce an acceleration? Explain.
\nSolution:<\/strong><\/span>
\nYes, the steering wheel can accelerate a car by changing its direction of motion.<\/p>\nChapter 6 Applications Of Newton’s Laws Q.19P<\/strong>
\nA certain spring has a force constant k. (a) If this spring is cut in half, does the resulting half spring have a force constant that is greater than, less than, or equal to k? (b) If two of the original full-length springs are connected end to end, does the resulting double spring have a force constant that is greater than, less than, or equal to k?
\nSolution:<\/strong><\/span>
\n
\n<\/p>\nChapter 6 Applications Of Newton’s Laws Q.20CQ<\/strong>
\n
\nSolution:<\/strong><\/span>
\nPeople on the outer rim of a rotating space station must experience a force directed toward the center of the station in order to follow a circular path. This force is applied by the floor of the station, which is really its outermost wall. Because people feel the upward force acting on them from the floor, just as they would on Earth, the sensation is like an artificial gravity.<\/p>\nChapter 6 Applications Of Newton’s Laws Q.20P<\/strong>
\nPulling up on a rope, you lift a 4.35-kg bucket of water from a well with an acceleration of 1.78 m\/s2. What is the tension in the rope?
\nSolution:<\/strong><\/span>
\n<\/p>\nChapter 6 Applications Of Newton’s Laws Q.21CQ<\/strong>
\nWhen rounding a corner on a bicycle or a motorcycle, the driver leans inward, toward the center of the circle. Why?
\nSolution:<\/strong><\/span>
\nWhen a bicycle rider leans inward on a turn, the force applied to the bicycle wheels by the ground is both upward and inward. It is this inward force that produces the rider\u2019s centripetal acceleration.<\/p>\nChapter 6 Applications Of Newton’s Laws Q.21P<\/strong>
\nWhen a 9.09-kg mass is placed on top of a vertical spring, the spring compresses 4.18 cm. Find the force constant of the spring.
\nSolution:<\/strong><\/span>
\n<\/p>\nChapter 6 Applications Of Newton’s Laws Q.22CQ<\/strong>
\nIn Robin Hood: Prince of Thieves, starring Kevin Costner, Robin swings between trees on a vine that is on fire. At the lowest point of his swing, the vine bums through and Robin begins to fall. The next shot, from high up in the trees, shows Robin falling straight downward. Would you rate the physics of this scene \u201cGood,\u201d \u201cBad,\u201d or \u201cUgly\u201d? Explain.
\nPROBLEMS AND CONCEPTUAL EXERCISES
\nNote: Answers to odd-numbered Problems and Conceptual Exercises can be found in the back of the book. IP denotes an integrated problem, with both conceptual and numerical parts; BIO identifies problems of biological or medical interest; CE indicates a conceptual exercise, Predict\/Explain problems ask for two responses: (a) your prediction of a physical outcome, and (b) the best explanation among three provided. On all problems, red bullets (,,) are used to indicate the level of difficulty.
\nSECTION 6-1 FRICTIONAL FORCES
\nSolution:<\/strong><\/span>
\nThe physics of this scene is somewhere between bad and ugly. When the rope burns through, the robin is moving horizontally. This horizontal motion should continue as the robin falls, leading to a parabolic trajectory rather than the straight downward drop shown in the movie.<\/p>\nChapter 6 Applications Of Newton’s Laws Q.22P<\/strong>
\nA 110-kg box is loaded into the trunk of a car. If the height of the car\u2019s bumper decreases by 13 cm, what is the force constant of its rear suspension?
\nSolution:<\/strong><\/span>
\n<\/p>\nChapter 6 Applications Of Newton’s Laws Q.23P<\/strong>
\nA 50.0-kg person takes a nap in a backyard hammock. Both ropes supporting the hammock are at an angle of 15.0\u00b0 above the horizontal, Find the tension in the ropes.
\nSolution:<\/strong><\/span>
\n<\/p>\nChapter 6 Applications Of Newton’s Laws Q.24P<\/strong>
\n
\nSolution:<\/strong><\/span>
\n<\/p>\nChapter 6 Applications Of Newton’s Laws Q.25P<\/strong>
\n
\nSolution:<\/strong><\/span>
\n<\/p>\nChapter 6 Applications Of Newton’s Laws Q.26P<\/strong>
\nThe equilibrium length of a certain spring with a force constant of k = 250 N\/m is 0.18 m. (a) What is the magnitude of the force that is required to hold this spring at twice its equilibrium length? (b) Is the magnitude of the force required to keep the spring compressed to half its equilibrium length greater than, less than, or equal to the force found in part (a)? Explain.
\nSolution:<\/strong><\/span>
\n
\n
\n<\/p>\nChapter 6 Applications Of Newton’s Laws Q.27P<\/strong>
\nIllinois Jones is being pulled from a snake pit with a rope that breaks if the tension in it exceeds 755 N. (a) If Illinois Jones has a mass of 70.0 kg and the snake pit is 3.40 m deep, what is the minimum tune that is required to pull our intrepid explorer from the pit? (b) Explain why the rope breaks if Jones is polled from the pit in less time than that calculated in part (a).
\nSolution:<\/strong><\/span>
\n
\n<\/p>\nChapter 6 Applications Of Newton’s Laws Q.28P<\/strong>
\n
\nSolution:<\/strong><\/span>
\n
\n
\n<\/p>\nChapter 6 Applications Of Newton’s Laws Q.29P<\/strong>
\nYour friend\u2019s 13.6-g graduation tassel hangs from his rearview mirror. (a) When he acceleration stoplight, the tassel deflects backward toward the car. Explain. (b) If the tassel hangs at an angle of 6.4 the vertical, what is the acceleration of the car?
\nSolution:<\/strong><\/span>
\n
\n
\n<\/p>\nChapter 6 Applications Of Newton’s Laws Q.30P<\/strong>
\nIn Problem 29, (a) find the tension in the siring holding the tassel. (b) At what angle to the vertical will the tension in the string be twice the weight of the tassel?
\nProblem 29
\nYour friend\u2019s 13.6-g graduation tassel hangs on a string from his rearview mirror. (a) When he accelerates from a stoplight, the tassel deflects backward toward the rear of the car. Explain. (b) If the tassel hangs at an angle of 6.44\u00b0 relative to the vertical, what is the acceleration of the car?
\nSolution:<\/strong><\/span>
\n
\n<\/p>\nChapter 6 Applications Of Newton’s Laws Q.31P<\/strong>
\n
\nSolution:<\/strong><\/span>
\n
\n
\n
\n<\/p>\nChapter 6 Applications Of Newton’s Laws Q.32P<\/strong>
\n
\nSolution:<\/strong><\/span>
\n
\n
\n<\/p>\nChapter 6 Applications Of Newton’s Laws Q.33P<\/strong>
\n
\nSolution:<\/strong><\/span>
\n
\n<\/p>\nChapter 6 Applications Of Newton’s Laws Q.34P<\/strong>
\nPulling the string on a bow back with a force of 28.7 lb, an archer prepares to shoot an arrow. If the archer pulls in the cen\u00adter of the string, and the angle between the two halves is 138\u00b0, what is the tension in the string?
\nSolution:<\/strong><\/span>
\n
\n<\/p>\nChapter 6 Applications Of Newton’s Laws Q.35P<\/strong>
\n
\nSolution:<\/strong><\/span>
\n
\n<\/p>\nChapter 6 Applications Of Newton’s Laws Q.36P<\/strong>
\n
\nSolution:<\/strong><\/span>
\n
\n<\/p>\nChapter 6 Applications Of Newton’s Laws Q.37P<\/strong>
\n
\nSolution:<\/strong><\/span>
\n
\n<\/p>\nChapter 6 Applications Of Newton’s Laws Q.38P<\/strong>
\n
\n
\nSolution:<\/strong><\/span>
\n
\n
\n
\n
\n
\n<\/p>\nChapter 6 Applications Of Newton’s Laws Q.39P<\/strong>
\n
\nSolution:<\/strong><\/span>
\n
\n<\/p>\nChapter 6 Applications Of Newton’s Laws Q.40P<\/strong>
\n
\nSolution:<\/strong><\/span>
\n
\n
\n<\/p>\nChapter 6 Applications Of Newton’s Laws Q.41P<\/strong>
\n
\n