Physics 11 Assignment KEY Dynamics Chapters 4 & 5


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1 Physics Assignment KEY Dynamics Chapters 4 & 5 ote: for all dynamics problemsolving questions, draw appropriate free body diagrams and use the aforementioned problemsolving method.. Define the following terms: Inertia  the natural tendency of an object to stay at rest or continue its motion in a straight line at constant speed in the absence of outside forces; objects with greater mass have greater inertia Dynamics  the study of the motions of bodies while considering their masses and the responsible forces Mechanics  the branch of physics comprising kinematics and dynamics; simply, the how and the why of simple motion ewton s Laws of Motion  three fundamental laws of motion which are the basis of ewtonian mechanics are: ) an object will remain at rest or in straightline motion unless acted on by an outside force; ) the acceleration of an object is proportional to the force acting on it and inversely proportional to its mass; 3) for every action force on an object, the object exerts and equal and opposite reaction force Force  an action, like a push or a pull, that causes a change in motion of an object Inertial frame of reference  a frame of reference in which the law of inertia is valid; it is a nonaccelerating frame of reference oninertial frame of reference  an accelerating frame of reference Mass  the quantity of matter an object contains; determined through the inertial properties of an object or its gravitational influence on other objects Inertial mass  the property of matter that resists a change in motion Gravitational mass  the property of matter that determines the strength of the gravitational force Contact force  the force exerted by an object in direct contact with another object oncontact force  the force that acts even though objects are separated by a distance, such as attraction or repulsion between magnets Weight  the force that gravity exerts on an object because of its mass Static frictional force  a frictional force that acts to keep an object at rest; measured as the force required to move an object from rest Kinetic frictional force  a frictional force that acts to slow the motion of an object; measured as the force required to just keep an object sliding over another object Coefficient of friction  the ratio of frictional force to the normal force between two object surfaces ormal force  a force that acts in a direction perpendicular to the common contact surface between two objects et force  the vector sum of all forces acting on an object. Why does a child in a wagon seem to fall backward when you give the wagon a sharp pull? The child tends to remain at rest (ewton s st Law), unless a force acts on her. The force is applied to the wagon, not the child, and so the wagon accelerates out from under the child, making it look like the child falls backwards relative to the wagon. If the child is standing in the wagon, the force of friction between the child and the bottom of the wagon will produce an acceleration of the feet, pulling the feet out from under the child, also making the child fall backwards.
2 3. If the acceleration of a body is zero, are no forces acting on it? If the acceleration of an object is zero, then by ewton s second law, the net force must be zero. There can be forces acting on the object as long as the vector sum of the forces is zero. 4. Why do you push harder on the pedals of a bicycle when first starting out than when moving at a constant speed? You push harder on the pedals of a bicycle when first starting out than when moving at constant velocity because your applied force must overcome static friction. Static friction is greater than kinetic friction. Once the bicycle is moving a constant velocity, less applied force is required to equal kinetic friction. 5. Only one force acts on an object. Can the object have zero acceleration? Can it have zero velocity? If only one force act on an object, then the net force is greater than zero and, according to the nd law of motion, the acceleration cannot be zero and it cannot have zero velocity. 6. The force of gravity on a kg rock is twice as great as that on a kg rock. Why then doesn t the heavier rock fall faster? The acceleration of both rocks is found by dividing their weight (the force of gravity on them) by their mass. The kg rock has a force of gravity on it that is twice as great as the force of gravity on the kg rock, but also twice as great a mass as the kg rock, so the acceleration is the same for both. 7. A person exerts an upward force of 40 to hold a bag of groceries. Describe the reaction force by stating (a) its magnitude, (b) its direction, (c) on what body it is exerted, and (d) by what body it is exerted. (a) The magnitude is 40. (b) The direction is downward. (c) It is exerted on the person. (d) It is exerted by the bag of groceries. 8. Why is the stopping distance of a truck much shorter than for a train going the same speed? The stopping distance for a truck is much shorter than that of a train going the same speed because, even though the rate of slowing down, i.e. acceleration, is the same, the truck s mass is significantly less than the train s mass. Therefore, according to the Second law of motion, a greater force will be required to slow down the train and thus a longer period of time and greater stopping distance. 9. You can hold a heavy box against a rough wall and prevent it from slipping down by pressing only horizontally. How can the application of a horizontal force keep an object from moving vertically? By pressing the block against the rough wall, you increase the normal force of the wall on the block. As the normal force increases, the amount of static friction between the block and the wall increases. This static friction force is vertical and opposes gravity or the
3 weight of the block. Provided that the static friction force equals the force of weight, the block will not slide down the wall. 0. a) A box sits at rest on a rough 30 o inclined plane. Draw free body diagram, showing all the forces acting on the box. b) How does the diagram change if the box were sliding down the plane? c) How does the diagram change if the box were sliding up the plane? (a) (b) (c) In (a) the friction is static and opposes the impending motion down the plane. In (b) the friction is kinetic and opposes the motion down the plane. In (c) the friction is kinetic and opposes the motion up the plane.. What force is needed to accelerate a child on a sled (total mass = 60.0 kg) at.5 m/s? The net force, F net, is due to some applied force, F a, and friction, F f. Since we do not know any of these individual forces, solve for Fnet. Fnet = m a = (60.0 kg)(.5 m/s ) = A net force of 55 accelerates a bike and rider at.0 m/s. What is the mass of the bike and rider? The net force, F net, is due to some applied force, F a, and friction, F f and is 55. F net = m a = F m a net = 55.0 m/s = 6 kg 3. How much tension must a rope withstand if it is used to accelerate a 050kg car horizontally at.0 m/s? Ignore friction.
4 The net force, F net, is due to tension in the rope, F T. F net = F = m a = (050 kg)(.0 m/s ) = 60 T 4. What is the weight of a 66kg astronaut (a) on Earth, (b) on the Moon (g =.7 m/s ), (c) on Mars (g = 3.7 m/s ), and (d) in outer space traveling with constant velocity? The mass remains the same in all four situations, 66 kg, but the weight is different: (a) = m g = (66 kg)(9.8m/s ) = 647 F G (b) F G (c) F G (d) F G = m g = (66 kg)(.7 m/s ) = = m g = (66 kg)(3.7 m/s ) = 44 = m g = (66 kg)(0 m/s ) = 0 5. A 0.0kg box rest on a table. (a) What is the weight of the box and the normal force acting on it? (b) A 0.0kg box is placed on top of the 0.0kg box. Determine the normal force that the table exerts on the 0.0kg box and the normal force that the 0.0kg box exerts on the 0.0kg box. (a) The weight of the box is: FG = m g = (0.0 kg)(9.8m/s ) = 96 The normal force acting on it is: F = F = m g = (0.0 kg)(9.8m/s ) = 96 G (b) We select both objects and apply the Second Law of Motion: F = F = ( m + m ) g = (0.0 kg kg)(9.8m/s ) = 94 G 0 0 If we select the top block as the object, then: F = F = m g = (0.0 kg)(9.8m/s ) = 98.0 G 6. What average force is required to stop an 00kg car in 8.0 s if it is traveling at 90 km/h?
5 The average force required to stop the car would be a combination of the braking force and the frictional force (call it F A ). The acceleration can be found from the car s D motion: v = v + at f i v f vi 0 m/s  5 m/s a = = = 3.3 m/s t 8.0 s We apply the Second Law of Motion: F A = m a = (00 kg)( 3.3 m/s ) = What average force is needed to accelerate a 7.00g pellet from rest to 75 m/s over a distance of m along the barrel of a rifle? The average force required to accelerate Bullet Bill comes from the explosion in the gun (call it F A ). The acceleration can be found: 8. A 0.0kg bucket is lowered by a rope in which there is 63.0 of tension. What is the acceleration of the bucket? Is it up or down? F = F F = m g F = m a net G T T (0 kg)(9.8 m/s ) (63 ) = (0 kg)(a) a = 3.5 m/s ( down) 9. A person stands on a bathroom scale in a motionless elevator. When the elevator begins to move, the scale briefly reads only 0.75 of the person s weight. Calculate the acceleration of the elevator and find the direction of acceleration. The scale reads the force the person exerts on the scale. From the Third Law of Motion, this is also the magnitude of the normal force acting on the person. F F = m a G m g F = m a mg 0.75mg = ma a = = (0.75)(9.8 m/s ).45 m/s ( down)
6 0. If the coefficient of kinetic friction between a 35kg crate and the floor is 0.30, what horizontal force is required to move the crate at a steady speed across the floor? What horizontal force is required if the coefficient of kinetic friction is zero? Because the crate is moving at constant speed, the acceleration is zero. Using the Second Law of Motion for each bucket: F = F  F = 0 net A f F = F = µ F = A f k (0.30)(35 kg)(9.8 m/s )=03 If µ k is reduced to zero then the friction is eliminated and no F A is required to keep the crate moving at constant speed (i.e. First Law of Motion). A force of 40.0 is required to start a 5.0kg box moving across a horizontal concrete floor. (a) What is the coefficient of static friction between the box and the floor? (b) If the force continues, the box accelerates at 0.70 m/s. What is the coefficient of kinetic friction? At the moment the box begins to move, the applied force equals the frictional force and acceleration is instantaneously zero: (a) F = F  F = 0 net A f F = F = µ F = 40.0 A f s Ff 40.0 µ s = = = 0.85 F (5.0 kg)(9.8 m/s ) (b) F = F  F = m a = (5.0 kg)(0.70 m/s ) = 3.5 net A f F = F 3.5 = = 36.5 f A Ff 36.5 µ s = = = F (5.0 kg)(9.8 m/s ). An Atwood machine consists of masses 3.8 kg and 4. kg. What is the acceleration of the masses? What is the tension in the rope? m a = g m m + m = 4. kg 3.8 kg 4. kg kg ( 9.8 m/s ) = 0.49 m/s 3. The smaller mass on an Atwood machine is 5. kg. If the masses accelerate at 4.6 m/s, what is the mass of the second object? What is the tension in the rope?
7 m a = g m a ( m + m ) = g( m m ) am am m m m + am gm m + m ( a g) = m ( a g) = gm a g = m a g = 4.4 kg = am gm gm 4.6 m/s 9.8 m/s = 5. kg 4.6 m/s 9.8 m/s 4. Stacie, who has a mass of 45 kg, starts down a slide that is inclined at an angle of 45 with the horizontal. If the coefficient of kinetic friction between Stacie s shorts and the slide is 0.5, what is her acceleration? 5. You are shadowing a nurse in the emergency room of a local hospital. An orderly wheels in a patient who has been in a very serious accident and has had severe bleeding. The nurse quickly explains to you that in a case like this, the patient s bed will be tilted with the head downward to make sure the brain gets enough blood. She tells you that, for most patients, the largest angle that the bed can be tilted without the patient beginning to slide off is 3.0 from the horizontal. a. On what factor or factors does this angle of tilting depend? The coefficient of static friction between the patient and the bed s sheets. b. Find the coefficient of static friction between a typical patient and the bed s sheets.
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