1403 Exam 2: Stanley
A ball of mass m is attached to a string of length L. It is being swung in a vertical circle with enough speed so that the string remains taut throughout the ball's motion. (Figure 1)Assume that the ball travels freely in this vertical circle with negligible loss of total mechanical energy. At the top and bottom of the vertical circle, the ball's speeds are vt and vb, and the corresponding tensions in the string are T t and T b. Tt and Tb have magnitudes Tt and Tb. Part A: Find Tb−Tt, the difference between the magnitude of the tension in the string at the bottom relative to that at the top of the circle.
First write the centripetal force eqs at top & bottom. Choose "in" toward the center as positive; At Top; Tt + mg = mvt^2/R At Bottom; Tb - mg = mvb^2/R Subtract Tb - Tt = 2mg + (1/R)[mvb^2 - mvt^2] Rewrite the last term using conservation of energy; (1/2)mvb^2 = (1/2)mvt^2 + 2mgR mvb^2 - mvt^2 = 4mgR Sub this into the tension eq; Tb - Tt = 2mg + 4mg = 6mg
IP A sled slides without friction down a small, ice-covered hill. On its first run down the hill, the sled starts from rest at the top of the hill and its speed at the bottom is 7.50 m/s. Part A: On a second run, the sled starts with a speed of 1.50 m/s at the top. What is the speed of the sled at the bottom of the hill on the second run?
Given that the speed of sled at bottom is v = 7.50 m/s --------------------------------------------------------- From the conservation of energy at top and bottom ofthe incline mgh = (1/2)mv2 h = v2 /2g = 2.869 m Let initial speed of the sled at height h is u = 1.50m/s From the conservation of energy at top and bottom ofthe incline mgh + (1/2)mu2 = (1/2)mv2 v = ( 2gh + u2 )1/2 = 7.65 m/s
A pendulum bob swings from point A to point B along the circular arc indicated in the figure. Part A: Indicate whether the work done on the bob by gravity is positive, negative, or zero. Part B: Indicate whether the work done on the bob by the string is positive, negative, or zero.
Part A: work done on the bob by gravity is positive. Part B: work done on the bob by the string is zero
A spring-loaded toy gun is used to shoot a ball of mass m=1.50kg straight up in the air, as shown in (Figure 1). The spring has spring constant k=667N/m. If the spring is compressed a distance of 25.0 centimeters from its equilibrium position y=0 and then released, the ball reaches a maximum height hmax (measured from the equilibrium position of the spring). There is no air resistance, and the ball never touches the inside of the gun. Assume that all movement occurs in a straight line up and down along the y axis. Part A: Which of the following statements are true?
Mechanical Energy is conserved because no dissipative forces perform work on the ball. The forces of gravity and the spring have potential eneriges associated with them
A spring-loaded toy gun is used to shoot a ball of mass m=1.50kg straight up in the air, as shown in (Figure 1). The spring has spring constant k=667N/m. If the spring is compressed a distance of 25.0 centimeters from its equilibrium position y=0 and then released, the ball reaches a maximum height hmax (measured from the equilibrium position of the spring). There is no air resistance, and the ball never touches the inside of the gun. Assume that all movement occurs in a straight line up and down along the y axis. Part A: Which of the following statements are true.
Mechanical energy is conserved because no dissipative forces perform work on the ball. The forces of gravity and the spring have potential energies associated with them
Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.40×106 N , one at an angle 17.0 ∘ west of north, and the other at an angle 17.0 ∘ east of north, as they pull the tanker a distance 0.690 km toward the north. Part A:What is the total work done by the two tugboats on the supertanker?
Part A:Find the net force acting on the object along displacement direction. The expression for the net force is,F= F1costheta+F2costheta After finding the Force plug it into W=Fd
A child sits in a wagon with a pile of 0.54-kg rocks Part A. If she can throw each rock with a speed of 9.5 m/s relative to the ground, causing the wagon to move, how many rocks must she throw per minute to maintain a constant average speed against a 3.4-N force of friction?
Part A:Ok, let's remember some equations Linear momentum = p = mass*speed = m*v Linear impulse = F*dt By Newton's second law : F = m*dv / dt dv / dt = acceleration dp / dt = m*dv / dt dp / dt = F >>> F*dt = dp >>> this equation we will use The variation of linear momentum is the linear impulse : for your problem, let's set n = number of rock n*(m*v) = F*(60s) n*(0.54*9.5) = F*60 If we want a constant speed = F = Fr = 3.1 N n*0.54*9.5 = 3.4*60 n = 40 aprox
A 0.28 kg pendulum bob is attached to a string 1.2 m long. What is the change in the gravitational potential energy of the system as the bob swings from point A to point B in the figure? Theta= 35
Since there is a decrease in potential energy when bob will move from A to B due to the decrease in height from an equilibrium position. The change in gravitational potential energy of the bob is given by: PE=-mgh (1-costheta)
A ball of mass m is attached to a string of length L. It is being swung in a vertical circle with enough speed so that the string remains taut throughout the ball's motion. (Figure 1)Assume that the ball travels freely in this vertical circle with negligible loss of total mechanical energy. At the top and bottom of the vertical circle, the ball's speeds are vt and vb, and the corresponding tensions in the string are Tt and Tb. Tt and Tb have magnitudes Tt and Tb. Part A: Find Tb−Tt, the difference between the magnitude of the tension in the string at the bottom relative to that at the top of the circle.
T(b) - mg = m * v(b)2 / L -------------> (1) T(t) + mg = m * v(t)2 / L -------------> (2) By conservation of energy, 1/2 * m * v(b)2 = [1/2 * m * v(t)2 ] + 2mgL v(b)2 - v(t)2 = 4gL subtract (1) & (2), we get T(b) - T(t) = 2mg + (m / L)*[v(b)2 - v(t)2] =2mg + (m/L)*4gL =2mg + 4mg T(b) - T(t) = 6mg
To make a bounce pass, a player throws a 0.50-kg basketball toward the floor. The ball hits the floor with a speed of 6.0 m/s at an angle of 75 ∘ to the vertical. If the ball rebounds with the same speed and angle, what was the magnitude of the impulse delivered to it by the floor?
The impulse I = change in momentum = 2mvcosθ = 2(0.5kg) (6m/s) cos75 = 1.60kg.m/s
Learning Goal
To understand how to apply the law of conservation of energy to situations with and without nonconservative forces acting. The law of conservation of energy states the following: In an isolated system the total energy remains constant. If the objects within the system interact through gravitational and elastic forces only, then the total mechanical energy is conserved. The mechanical energy of a system is defined as the sum of kinetic energy K and potential energy U. For such systems in which no forces other than the gravitational and elastic forces do work, the law of conservation of energy can be written as Ki+Ui=Kf+Uf, where the quantities with subscript "i" refer to the "initial" moment and those with subscript "f" refer to the final moment. A wise choice of initial and final moments, which is not always obvious, may significantly simplify the solution. The kinetic energy of an object that has mass m and velocity v is given by K=12mv2. Potential energy, in contrast, has many forms. Two forms that you will be dealing with often are gravitational and elastic potential energy. Gravitational potential energy is the energy possessed by elevated objects. For small heights, it can be found as Ug=mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the elevation of the object above the zero level. The zero level is the elevation at which the gravitational potential energy is assumed to be zero. The choice of the zero level is dictated by convenience; typically (but not necessarily), it is selected to coincide with the lowest position of the object during the motion explored in the problem. Elastic potential energy is associated with stretched or compressed elastic objects such as springs. For a spring with a force constant k, stretched or compressed a distance x, the associated elastic potential energy is Ue=12kx2. When all three types of energy change, the law of conservation of energy for an object of mass m can be written as 12mv2i+mghi+12kx2i=12mv2f+mghf+12kx2f. The gravitational force and the elastic force are two examples of conservative forces. What if nonconservative forces, such as friction, also act within the system? In that case, the total mechanical energy will change. The law of conservation of energy is then written as 12mv2i+mghi+12kx2i+Wnc=12mv2f+mghf+12kx2f, where Wnc represents the work done by the nonconservative forces acting on the object between the initial and the final moments. The work Wnc is usually negative; that is, the nonconservative forces tend to decrease, or dissipate, the mechanical energy of the system
Two springs, with force constants k1 and k2, are connected in series, as shown in the figure. Part A: How much work is required to stretch this system a distance x from the equilibrium position?
W=1/2 ((k1k2)/(k1+k2)) (x^2)
The Center of Mass of Water Find the center of mass of a water molecule, referring to the figure(Figure 1) for the relevant angles and distances. The mass of a hydrogen atom is 1.0 u, and the mass of an oxygen atom is 16 u, where the atomic mass unit, u, is defined as follows: 1u=1.66×10−27kg.). Use the center of the oxygen atom as the origin of your coordinate system.
Xcm=m1x1+m2x2/m1+m2 2(1.0)(9.6x10^-11)cos(104.5)+(16)(0)/2(1.0)+16 =6.5x10^-12m,0
A 0.35-kg croquet ball is initially at rest on the grass. When the ball is struck by a mallet, the average force exerted on it is 200 N. If the ball's speed after being struck is 5.1 m/s, for what amount of time was the mallet in contact with the ball?
v= at F= ma: a= F/m = 200/.35= 571.429 m/s^2 t= v/1 = 5.1/ 571.429 = 8.9s
Momentum and kinetic energy are both functions of mass and velocity of the object. Momentum is directly proportional to the velocity and kinetic energy is proportional to the square of the velocity. While both momentum and kinetic energy are directly proportional to the mass. The expression for the momentum of an object is given as follows: Here, is the momentum, is the mass and is the velocity. The kinetic energy, mass, and velocity of a particle are related by the equation as follows: Here is the kinetic energy of the particle, is mass of the particle and is the velocity of the particle Using the expression for momentum and kinetic energy one can calculate the terms provided the variables are known for the objects under discussion and results can be compared. Consider two objects (Object 1 and Object 2) moving in the same direction on a frictionless surface. Object 1 moves with speed v1=v and has mass m1=2m. Object 2 moves with speed v2=2√v and has mass m2=m. A: Which object has the larger magnitude of its momentum? B: Which object has the larger kinetic energy?
A)Calculate the momentum of object 1 as follows; P1=m1v1 Substitute 2m for m1 and v for v1 the equation P1=m1v1 to calculate the momentum. P1=(2m)(V) = 2mv Calculate the momentum of object 2 as follows; P2=(m2)(v2) Substitute m for m2 and square root 2v in the equation P2= m2v2 to calculate the momentum. P2= square root 2vm B: Calculate the kinetic energies of both objects as follows: Substitute 2 v for in the equation KE1= 1/2 mv^2 to calculate the kinetic energy of object 1: KE1=m(2)^2 =2m Substitute 2 v for in the equation KE1= 1/2 mv^2 to calculate the kinetic energy of object 2: KE1=m(2)^2 =2m They have the same KE.
Learning Goal: Understand that conservative forces can be removed from the work integral by incorporating them into a new form of energy called potential energy that must be added to the kinetic energy to get the total mechanical energy. The first part of this problem contains short-answer questions that review the work-energy theorem. In the second part we introduce the concept of potential energy. But for now, please answer in terms of the work-energy theorem. Work-Energy Theorem The work-energy theorem states: Kf=Ki+Wall, where Wall is the work done by all forces that act on the object, and Ki and Kf are the initial and final kinetic energies, respectively. Part A: The work-energy theorem states that a force acting on a particle as it moves over a ______ changes the ______ energy of the particle if the force has a component parallel to the motion. Part B: To calculate the change in kinetic energy, you must know the force as a function of _______. The work done by the force causes the kinetic energy change. Part C: To illustrate the work-energy concept, consider the case of a stone falling from xi to xf under the influence of gravity. Using the work-energy concept, we say that work is done by the gravitational _____, resulting in an increase of the ______ energy of the stone. Part D: Rather than ascribing the increased kinetic energy of the stone to the work of gravity, we now (when using potential energy rather than work-energy) say that the increased kinetic energy comes from the ______ of the _______ energy. Part E: This process happens in such a way that total mechanical energy, equal to the ______ of the kinetic and potential energies, is _______.
A: Distance/ Kinetic: It is important that the force has a component acting in the direction of motion. For example, if a ball is attached to a string and whirled in uniform circular motion, the string does apply a force to the ball, but since the string's force is always perpendicular to the motion it does no work and cannot change the kinetic energy of the ball. B: Position C: Force/ Kinetic D: Change/ Potential E: Sum/ Zero
A roller-coaster car may be represented by a block of mass 50.0 kg . The car is released from rest at a height h = 45.0 m above the ground and slides along a frictionless track. The car encounters a loop of radius R = 15.0 m at ground level, as shown. As you will learn in the course of this problem, the initial height 45.0 m is great enough so that the car never loses contact with the track. Part A: Find the kinetic energy K of the car at the top of the loop. Part B: Find the minimum initial height hmin at which the car can be released that still allows the car to stay in contact with the track at the top of the loop.
A: Find an expression for the kinetic energy K of the car at the top of the loop. Express the kinetic energy numerically, in joules. Since the radius of the circle is 15.0m, the diameter is 30.0m. Therefore, at the top of the loop, the height of the car will be 30.0m. Solve for KE by first finding potential energy at the beginning of the track: U = mgh U = 50.0(9.8)(45.0) U = 22050J Since there is no friction, there will be a free exchange of potential to/from kinetic energy. So if we find the potential energy at the top of the loop and subtract that from our beginning potential energy, we'll have the kinetic energy: Utop = mgh Utop = 50.0(9.8)(30.0) Utop = 14700J KE = 22050 - 14700 KE = 7350J Part B: Find the minimum initial height hmin at which the car can be released that still allows the car to stay in contact with the track at the top of the loop. This is a centripetal force question. Solve for the velocity at the top of the loop using the formula: ac = V2/r At the top of the loop and minimum velocity, the centripetal force equals the force of gravity (and centripetal acceleration equals acceleration due to gravity). So: 9.8 = V2/r, 9.8 = V2/15, 147 = V2, V = 12.12m/s Now solve for KE at the top of the loop: KE = 1/2mv2 KE = 1/2(50.0)(12.12)2 KE = 3675J Next, solve for potential energy at the top of the loop: U = mgh U = 50.0(9.8)(30.0) U = 14700J Add the two energies: 3675 + 14700 = 18375 And now solve for height using the above as beginning potential energy: U = mgh 18375 = 50(9.8)(h) h = 37.5m
Human-Powered Flight Human-powered aircraft require a pilot to pedal, as in a bicycle, and produce a sustained power output of about 0.30 hp. The Gossamer Albatross flew across the English Channel on June 12, 1979, in 2 h 49 min. Part A: How much energy did the pilot expend during the flight? Part B: How many Snickers candy bars (280 Cal per bar) would the pilot have to consume to be "fueled up" for the flight? [Note: The nutritional calorie, 1 Cal, is equivalent to 1000 calories (1000 cal) as defined in physics. In addition, the conversion factor between calories and joules is as follows: 1 Cal = 1000 cal = 1 kcal = 4186 J.]
A: First step is to convert HP to Watts. Then convert 2h 49 mins to mins. And then just plug into W=Pt B: In order to find number of snickers n=E/280(4186)
As illustrated in the figure, a spring with spring constant k is stretched from x=0 to x=3d, where x=0 is the equilibrium position of the spring. Part A: During which interval is the largest amount of energy required to stretch the spring Part B: A spring is stretched from x=0 to x=dx=d, where x=0 is the equilibrium position of the spring. It is then compressed from x=0 to x=−d. What can be said about the energy required to stretch or compress the spring? Part C: Now consider two springs A and B that are attached to a wall. Spring A has a spring constant that is four times that of the spring constant of spring B. If the same amount of energy is required to stretch both springs, what can be said about the distance each spring is stretched? Part D: Two identical springs are attached to two different masses, MA and MB, where MA is greater than MB. The masses lie on a frictionless surface. Both springs are compressed the same distance, d, as shown in the figure. Which of the following statements descibes the energy required to compress spring A and spring B?
A: From x=2d to x=3d (Recall that on a graph of force as a function of position, the work done by the force is represented by the area under the curve. The work done by the hand in the first segment to pull the spring from x=0 to x=d is represented by a single triangle. The area under the second segment from x=dto x=2d is three times larger than the first segment, and the area under the third segment from x=2d to x=3d is five times larger than in the first segment. So more energy is required to pull the spring through the third segment. ) B: The same amount of energy is required to either stretch or compress the spring. ( The work done to stretch or compress a spring from equilibrium is given by Won spring=12kx2, where x is the distance away from equilibrium that the spring moves. Since x is squared in the equation for work, stretching (x>0) or compressing (x<0) a spring by the same distance requires the same positive amount of work. ) C: Spring A must stretch half the distance spring B stretches. (The energy required to stretch a spring is proportional to k and to x2. If kA is four times kB, xA must be half that of xB, so the energy required is the same for both springs.) D: Spring A requires more energy than spring B. (Good job; you have realized an important fact. The work done on a spring to compress it a distance d is given by 12kd2. The amount of mass attached to the spring does not affect the work required to stretch or compress the spring)
On a frictionless horizontal air table, puck A (with mass 0.245 kg ) is moving toward puck B (with mass 0.371 kg ), which is initially at rest. After the collision, puck A has velocity 0.118 m/s to the left, and puck B has velocity 0.654 m/s to the right. A: What was the speed vAi of puck A before the collision? B: Calculate ΔK, the change in the total kinetic energy of the system that occurs during the collision.
A: Law of conservation of momentum states that a collision occurs between two bodies the total momentum before the collision is equal to the total momentum after the collision. vi= (.245)(-.118) + (.371)(.654) / (.245) = .872 m/s B: 1/2 (.245)(.872)= .0966 1/2 (.245)(-.118)^2+ (.371)(.654) = .2443
Learning Goal: To understand how to apply the law of conservation of energy to situations with and without nonconservative forces acting. The law of conservation of energy states the following: In an isolated system the total energy remains constant. If the objects within the system interact through gravitational and elastic forces only, then the total mechanical energy is conserved. The mechanical energy of a system is defined as the sum of kinetic energy K and potential energy U. For such systems in which no forces other than the gravitational and elastic forces do work, the law of conservation of energy can be written as: Ki+Ui=Kf+Uf, where the quantities with subscript "i" refer to the "initial" moment and those with subscript "f" refer to the final moment. A wise choice of initial and final moments, which is not always obvious, may significantly simplify the solution. The kinetic energy of an object that has mass m and velocity v is given by K=12mv2. Potential energy, in contrast, has many forms. Two forms that you will be dealing with often are gravitational and elastic potential energy. Gravitational potential energy is the energy possessed by elevated objects. For small heights, it can be found as Ug=mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the elevation of the object above the zero level. The zero level is the elevation at which the gravitational potential energy is assumed to be zero. The choice of the zero level is dictated by convenience; typically (but not necessarily), it is selected to coincide with the lowest position of the object during the motion explored in the problem. Elastic potential energy is associated with stretched or compressed elastic objects such as springs. For a spring with a force constant k, stretched or compressed a distance x, the associated elastic potential energy is Ue=12kx2. When all three types of energy change, the law of conservation of energy for an object of mass m can be written as 12mv2i+mghi+12kx2i=12mv2f+mghf+12kx2f. The gravitational force and the elastic force are two examples of conservative forces. What if nonconservative forces, such as friction, also act within the system? In that case, the total mechanical energy will change. The law of conservation of energy is then written as 12mv2i+mghi+12kx2i+Wnc=12mv2f+mghf+12kx2f, where Wnc represents the work done by the nonconservative forces acting on the object between the initial and the final moments. The work Wnc is usually negative; that is, the nonconservative forces tend to decrease or dissipate, the mechanical energy of the system. Part A: Which word in the statement of this problem allows you to assume that the table is frictionless? Part B: Which form of the law of conservation of energy describes the motion of the block when it slides from the top of the table to the bottom of the ramp? Part C: As the block slides down the ramp, what happens to its kinetic energy K, potential energy U, and total mechanical energy E? Part D: Using conservation of energy, find the speed vb of the block at the bottom of the ramp. Part E: Which form of the law of conservation of energy describes the motion of the block as it slides on the floor from the bottom of the ramp to the moment it stops? Part F: As the block slides across the floor, what happens to its kinetic energy K, potential energy U, and total mechanical energy E? Part G: What force is responsible for the decrease in the mechanical energy of the block? Part H: Find the amount of energy E dissipated by friction by the time the block stops
A: Smooth B: 1/2mvi2+mghf=1/2mf2+mghf C: K increases, U decrease, E stays the same D: vb: v^2+2gh square root E: 1/2 mvi2+Wnc= 1/2mvf2 F: K decreases, U stays the same and decreases G: friction H: Find the amount of E dissipated by friction by the time the block stops.
The soles of a popular make of running shoe have a force constant of 2.0×105 N/m . Treat the soles as ideal springs for the following questions. Part A: If a 62 kg person stands in a pair of these shoes, with her weight distributed equally on both feet, how much does she compress the soles? Part B: How much energy is stored in the soles of her shoes when she's standing?
A: Standing in the shoes the forces are balanced and and oppositeso Fweight = Fspring .5mg =Fkx, .5mg because you aresplitting the weight between two shoes .5(62kg)(9.8m/s2) = (2.0 x 105 N/m) x 303.8N = (2.0 x 105 N/m) x x = .001519m B: Potential Elastic Energy PE = .5(Fk)x2 PE = .5(2.0 x 105)(.001519m)2 PE = .231J, This is per shoe so doubleit to get the total potential energy PE = (.231J) x 2 PE = .46J
The two toy cars shown in the figure, with masses as given in the figure, (Figure 1) are ready to race. Both cars begin from rest. For each question, state whether the correct answer is car A, car B, or whether the two cars have equal values for the parameter in question. A: Which car crosses the finish line 1.0 m away first? B: Which car has the larger kinetic energy when it crosses the finish line 1.0 m away? C: Which car has a larger momentum when it crosses the finish line 1.0 m away? D: Which car has traveled farther after 10 s? E: After 10 s which car has a larger kinetic energy? F: After 10 s which car has a larger momentum?
A: The speed is directly proportional to acceleration and time is inversely proportional to velocity. Therefore, the time is directly proportional to mass. The heavier car takes more time to cross the finish line. The car B crosses first as it is lighter than car A. B: The kinetic energy depends on the force applied and distance moved by the cars. The force applied and distance moved is same for both the cars and so is the kinetic energy. C: The momentum is directly proportional to mass of the object if they have same kinetic energies. So, using the relation in masses of the cars compare the momentum. D: Car B E: Car B F: Both cars have the same momentum
A cardinal (Richmondena cardinalis) of mass 4.30×10−2 kg and a baseball of mass 0.147 kg have the same kinetic energy. What is the ratio of the cardinal's magnitude pc of momentum to the magnitude pb of the baseball's momentum?
A:Since the kinetic energy is the same, start by solving for the relative velocities: KE = 1/2mv2 Cardinal: KE = 1/2mv2 KEc = 1/2(0.0430)vc2 KEc = 0.0215vc2 Baseball: KE = 1/2mv2 KEb = 0.0735vc2 Set equal to each other: KEc = KEb 0.0215vc2 = 0.0735vb2 vc2 = 3.41vb2 vc = 1.84vb So the velocity of the cardinal is 1.84 times that of the baseball. Now plug into the momentum equation: p = mv Cardinal: pc = mcvc pc = (0.043)(1.84) pc = 0.07912 Baseball: pb = mbvb pb = (0.147)(1) pb = 0.147 The '1′ for the baseball's velocity just comes from the relative velocities calculated earlier (the cardinal's velocity is 1.82 times greater, so 1.82 : 1 is the ratio for velocity). Set equal to solve for the ratio: pc = pb 0.07912 = 0.147 0.541 = 1 B: A man weighing 660 N and a woman weighing 490 N have the same momentum. What is the ratio of the man's kinetic energy Km to that of the woman Kw? First, calculate the masses (just divide weight by g): Man = 710/9.8 = 67.35kg Woman = 430/9.8 = 50kg And follow a similar procedure as Part A, but starting with relative momentum this time: Man:pman = mmvm pman = 67.35vm Woman: pwoman = mwvw pwoman = 50vw Now solve for relative velocity since the momentum's were given as equal: pman = pwoman 67.35vm = 50vw vm = 0.742vw Now solve for relative kinetic energy: KE = 1/2mv2 Man: KEm = 1/2mmvm2 KEm = 1/2(67.35)(0.742)2 KEm = 18.54J Woman: KEw = 1/2mwvw2 KEw = 1/2(50)(1)2 KEw = 25J Set equal to find relative kinetic energy: 18.54J = 25J 0.742 = 1
Two ice skaters stand at rest in the center of an ice rink. When they push off against one another the 63-kg skater acquires a speed of 0.61 m/s Part A. If the speed of the other skater is 0.88 m/s , what is this skater's mass?
Part A: Use M1V1=M2V2 M2=M1V1/V2 63(.61)/(.88)=44
A 0.454-kg block is attached to a horizontal spring that is at its equilibrium length, and whose force constant is 22.0 N/m. The block rests on a frictionless surface. A 5.50×10−2-kg wad of putty is thrown horizontally at the block, hitting it with a speed of 8.99 m/s and sticking. A. How far does the putty-block system compress the spring?
Part A: mv1=m+M(v) v= (.050)(8.99)/(.050+.454) x= square root of m+M(v)^2/k
An ice cube is placed in a microwave oven. Suppose the oven delivers 125 W of power to the ice cube and that it takes 33200 J to melt it. Part A: How long does it take for the ice cube to melt?
Part A: In order to find you take the total joule/total W. Then divide by 60 seconds.
A young hockey player stands at rest on the ice holding a 1.1-kg helmet. The player tosses the helmet with a speed of 6.4 m/s in a direction 12 ∘ above the horizontal, and recoils with a speed of 0.30 m/s. A: Find the mass of the hockey player.
Part A: M2= m1v1costheata/v2
A honeybee with a mass of 0.150 g lands on one end of a floating 4.75-g popsicle stick, as shown in (Figure 1). After sitting at rest for a moment, it runs toward the other end with a velocity vb relative to the still water. The stick moves in the opposite direction with a speed of 0.100 cm/s Part A. What is the velocity of the bee? (Let the direction of the bee's motion be the positive x direction.)
Part A: Momentum is conserved m1v1=m2v2 (.15)(v1)=(4.75)(0.001) v1 =.0317 = 3.17
The recoil of a shotgun can be significant. Suppose a 3.6-kg shotgun is held tightly by an arm and shoulder with a combined mass of 15.0 kg. Part A. When the gun fires a projectile with a mass of 0.037 kg and a speed of 390 m/s, what is the recoil velocity of the shotgun and arm-shoulder combination?
Part A: Given data: Mass of shotgun: m1= 3.6 kg Mass of arm and shoulder m2= 15kg Mass of projectile m=.037 Speed of projectile v= 390 m/s total mass of shotgun and arm shoulder: M= m1+m2 = 18.6kg Calculate the recoio velocity of shotgun using law of conservation of momentum MV+mv=0 V=-mv/M
Olaf is standing on a sheet of ice that covers the football stadium parking lot in Buffalo, New York; there is negligible friction between his feet and the ice. A friend throws Olaf a ball of mass 0.400 kg that is traveling horizontally at 10.8 m/s . Olaf's mass is 74.1 kg .(Figure 1) a. If Olaf catches the ball, with what speed vf do Olaf and the ball move afterward? b. If the ball hits Olaf and bounces off his chest horizontally at 8.40 m/s in the opposite direction, what is his speed vf after the collision?
If Olaf catches the ball, with what speed vf do Olaf and the ball move afterward? Express your answer numerically in centimeters per second. Remember that momentum is conserved, so the momentum before and after Olaf catches the ball will be the same. Before Olaf catches the ball, the momentum is (mass * velocity, or 0.400 * 10.8) 4.68 kg·m/s. This won't change after Olaf catches the ball, so we can solve for the velocity: ρ = m * v Part B:If the ball hits Olaf and bounces off his chest horizontally at 7.40 m/s in the opposite direction, what is his speed after the collision? Express your answer numerically in centimeters per second. Remember momentum is conserved. In Part A we saw that the total momentum is (mass * velocity, or 0.400 * 11.7) 4.68 kg·m/s. This will stay the same after the collision. Since the ball will be moving backwards (negative) at 7.40 m/s, it's momentum after the collision is (7.40 * 0.400) 2.96 kg·m/s. So we can solve for Olaf's momentum: ρOlaf = 4.68 - ρball ρOlaf = 4.68 - (- 2.96) ρOlaf = 7.64 Now we can solve for Olaf's velocity: ρ = m * v 7.64 = 69.2 * v v = 0.110 m/s For some reason, Mastering Physics wants this in centimeters per second, so just multiply by 100. The answer is 11.0 cm/s 11.0 cm/s
A 1.00-kg block of wood sits at the edge of a table, 0.780 m above the floor. A 1.20×10−2-kg bullet moving horizontally with a speed of 715 m/s embeds itself within the block. Part A. What horizontal distance does the block cover before hitting the ground?
Part A: The mass of block M = 1 kg The mass of bullet m = 0.0120 kg height h = 0.780 m speed of bullet v = 715 m according to conservation of energy m v + M v' = ( M + m ) V The initial speed of the block v ' = 0 m/s So that we have m v =( M + m ) V ==> therefore final speed of the velocity - blocksystem is V = m v / ( M + m) = 0.0120 * 715 / ( 1 + 0.0120) = 8.478 m/s The time taken to hit the ground is t = √2h / g = √2 * 0.780 / 9.8 = 0.399 s Therefore the horizontal distance traveled by the block = V * t = 8.478 * 0.399 = 3.38 m
In this problem, you will calculate the location of the center of mass for the Earth-Moon system, and then you will calculate the center of mass of the Earth-Moon-Sun system. The mass of the Moon is 7.35×1022 kg , the mass of the Earth is 6.00×1024 kg , and the mass of the sun is 2.00×1030 kg . The distance between the Moon and the Earth is 3.80×105 km . The distance between the Earth and the Sun is 1.50×108 km. Part A: Calculate the location xcm of the center of mass of the Earth-Moon system. Use a coordinate system in which the center of the Earth is at x=0 and the Moon is located in the positive x direction. Part B:Where is the center of mass of the Earth-Moon system? The radius of the Earth is 6378 km and the radius of the Moon is 1737 km. Select one of the answers below: Part C: Calculate the center of mass for the sun-earth-moon system
Part A: xcm= m1x1+m2x2 total mass of the system is: M= m1+m2 (6.00x10^24kg)+(7.35x10^22kg) = 6.0735x10^24 now xcm= 6.0735x10^24kg(0m)+ (7.35x10^22kg)(3.80x10^5km) = 4.60x10^3km Part B:In this case, the Earth is the center of the coordinate system. Hence the distance between the origin and the Earth is zero. Thus the center of mass of the system depends upon the mass and position of the Moon and the total mass of the system. Part C: xcm= mexo+(mm)(xm)+(ms)(xs)/ mm+me+ms (6.0735x10^24kg)(0km)+(7.35x10^22kg)(3.80x10^5km)+(2.00x10^30kg)(1.50x10^8km)/(6.0735x10^24kg_+(7.35x10^22kg)+(2.00x10^30kg)
Three balls are thrown upward with the same initial speed v0,but at different angles relative to the horizontal, as shown in the figure. Part A: Ignoring air resistance, indicate which of the following statements is correct:
Part A: At the dashed level all three balls have the same speed.
Learning Goal: Understand that conservative forces can be removed from the work integral by incorporating them into a new form of energy called potential energy that must be added to the kinetic energy to get the total mechanical energy. The first part of this problem contains short-answer questions that review the work-energy theorem. In the second part, we introduce the concept of potential energy. But for now, please answer in terms of the work-energy theorem. Part A: Work-Energy Theorem The work-energy theorem states Kf=Ki+Wall, where Wall is the work done by all forces that act on the object, and Ki and Kf are the initial and final kinetic energies, respectively. Part A Part complete The work-energy theorem states that a force acting on a particle as it moves over a ______ changes the ______ energy of the particle if the force has a component parallel to the motion. Part B: To calculate the change in kinetic energy, you must know the force as a function of _______. The work done by the force causes the kinetic energy change. Part C: To illustrate the work-energy concept, consider the case of a stone falling from xi to xf under the influence of gravity. Using the work-energy concept, we say that work is done by the gravitational _____, resulting in an increase of the ______ energy of the stone. Part D: Rather than ascribing the increased kinetic energy of the stone to the work of gravity, we now (when using potential energy rather than work-energy) say that the increased kinetic energy comes from the ______ of the _______ energy. Part E: This process happens in such a way that total mechanical energy, equal to the ______ of the kinetic and potential energies, is _______.
Part A: Distance/Kinetic Part B: Position Part C: Force/ Kinetic Part D: Change/Potential Part E: Sum/ Conserved
A roller-coaster car may be represented by a block of mass 50.0 kg . The car is released from rest at a height h = 45.0 m above the ground and slides along a frictionless track. The car encounters a loop of radius R = 15.0 m at ground level, as shown. As you will learn in the course of this problem, the initial height 45.0 m is great enough so that the car never loses contact with the track. Part A: Find the kinetic energy K of the car at the top of the loop. Part B: Find the minimum initial height hmin at which the car can be released that still allows the car to stay in contact with the track at the top of the loop.
Part A: Find an expression for the kinetic energy K of the car at the top of the loop. Express the kinetic energy numerically, in joules. Since the radius of the circle is 15.0m, the diameter is 30.0m. Therefore, at the top of the loop, the height of the car will be 30.0m. Solve for KE by first finding potential energy at the beginning of the track: U = mgh U = 50.0(9.8)(45.0) U = 22050J Since there is no friction, there will be a free exchange of potential to/from kinetic energy. So if we find the potential energy at the top of the loop and subtract that from our beginning potential energy, we'll have the kinetic energy: Utop = mgh Utop = 50.0(9.8)(30.0) Utop = 14700J KE = 22050 - 14700= 7370J Part B: Find the minimum initial height hmin at which the car can be released that still allows the car to stay in contact with the track at the top of the loop. This is a centripetal force question. Solve for the velocity at the top of the loop using the formula: ac = V2/r At the top of the loop and minimum velocity, the centripetal force equals the force of gravity (and centripetal acceleration equals acceleration due to gravity). So: 9.8 = V2/r 9.8 = V2/15 147 = V2 V = 12.12m/s Now solve for KE at the top of the loop: KE = 1/2mv2 KE = 1/2(50.0)(12.12)2 KE = 3675J Next, solve for potential energy at the top of the loop: U = mgh U = 50.0(9.8)(30.0) U = 14700J Add the two energies: 3675 + 14700 = 18375 And now solve for height using the above as beginning potential energy: U = mgh 18375 = 50(9.8)(h) h = 37.5m
As illustrated in the figure, a spring with spring constant k is stretched from x=0 to x=3d, where x=0 is the equilibrium position of the spring Part A: During which interval is the largest amount of energy required to stretch the spring? Part B: A spring is stretched from x=0 to x=d, where x=0 is the equilibrium position of the spring. It is then compressed from x=0 to x=−d. What can be said about the energy required to stretch or compress the spring? Part C: Now consider two springs A and B that are attached to a wall. Spring A has a spring constant that is four times that of the spring constant of spring B. If the same amount of energy is required to stretch both springs, what can be said about the distance each spring is stretched? Part D: Two identical springs are attached to two different masses, MA and MB, where MA is greater than MB. The masses lie on a frictionless surface. Both springs are compressed the same distance, d, as shown in the figure. Which of the following statements descibes the energy required to compress spring A and spring B?
Part A: From x=2d to x=3 Part B: The same amount of energy is required to either stretch or compress the spring Part C: Spring A must stretch 2 times as far as spring B. Part D: Spring A requires less energy than spring B.
The human small intestine moves food along its 6.9-m length by means of peristalsis, a contraction and relaxation of muscles that propagate in a wave along the digestive tract. Part A: If the average force exerted by peristalsis is 0.24 N, how much work does the small intestine do to move food along its entire length?
Part A: In order to find W use W=Fd
A 0.590-kg bucket rests on a scale. Into this bucket you pour sand at the constant rate of 85.0 g/s. Part A. If the sand lands in the bucket with a speed of 4.30 m/s , what is the reading of the scale when there is 0.720 kg of sand in the bucket? Part B. What is the weight of the bucket and the 0.720 kg of sand?
Part A: Let d be the Greek letter delta (for change) F = d p / d t = change in momentum / time F = v dm / dt change in momentum due to 54 g/s accumulating in bucket F = 3.2 *. 590= 2.537 N Weight of sand and bucket = (.85 + .720) * 9.8 = 13.2 N Reading on scale = 13.2 + /2.537 = 14.8 N
Small birds like that in (Figure 1) can migrate over long distances without feeding, storing energy mostly as fat rather than carbohydrate. Fat is a good form of energy storage because it provides the most energy per unit mass: 1.00 grams of fat provides about 9.40 (food) Calories, compared to 4.20 (food) Calories per 1.00 grams of carbohydrate. Remember that Calories associated with food, which are always capitalized, are not exactly the same as calories used in physics or chemistry, even though they have the same name. More specifically, one food Calorie is equal to 1000 calories of mechanical work or 4184 joules. Therefore, in this problem use the conversion factor A: Consider a bird that flies at an average speed of 10.7 m/s and releases energy from its body fat reserves at an average rate of 3.70 W (this rate represents the power consumption of the bird). Assume that the bird consumes 4.00g of fat to fly over a distance db without stopping for feeding. How far will the bird fly before feeding again? B: How many grams of carbohydrate mcarb would the bird have to consume to travel the same distance db? C: Field observations suggest that a migrating ruby-throated hummingbird can fly across the Gulf of Mexico on a nonstop flight traveling a distance of about 800 km . Assuming that the bird has an average speed of 40.0 km/hr and an average power consumption of 1.70 W , how many grams of fat mfat does a ruby-throated hummingbird need to accomplish the nonstop flight across the Gulf of Mexico?
Part A: The problem tells us that one gram of fat has 9.4 Calories. Each Calorie equals 4186 J, so that means a gram of fat contains about 39,348 J of energy, and 4 grams of fat has about 157,394 J of energy. One Watt is equal to 1 joule per second, so a bird that burns energy at a rate of 3.7 W is burning 3.7 joules a second. This means it can fly for about 42,539 seconds (157,394 / 3.7). At a speed of 10.7 m/s, it will fly for about 455,165 m, which equals 455.165 km. Part B: The problem tells you that one gram of fat has about 9.4 Calories of energy, compared to 4.2 for a gram of carbs. So there is 2.2381 times as much energy in a gram of fat. If the bird consumes 4 grams of fat during its flight, it would have to consume 2.2381 * 4 = 8.9523 grams of carbs. Part C: To fly 800 km will take about 20 hours (at 40km/hr). This is equal to 72,000 seconds. At a power consumption of 1.7 W, the bird will consume (1.7 * 72,000) 122,400 J of energy. Since a gram of fat contains about (9.4 Calories * 4186 J/Calorie) 39348 joules of energy, the bird will need (122,400 / 39,348) about 3.1107 grams of fat.
A young woman on a skateboard is pulled by a rope attached to a bicycle. The velocity of the skateboarder is v= (3.6 m/s )x^ and the force exerted on her by the rope is F = (17 N)x^ + (12 N)y^. Part A: Find the work done on the skateboarder by the rope in 20 seconds. Part B: Assuming the velocity of the bike is the same as that of the skateboarder, find the work the rope does on the bicycle in 20 seconds.
Part A: There are no forces in the y-direction, only in the x direction. F=N(x)+N(y)= F-Pt Part B:Wb=-Ws A is positive because the force is in the same direction as v B is negative because the force is opposite of v
A hoop of mass M and radius R rests on a smooth, level surface. The inside of the hoop has ridges on either side, so that it forms a track on which a ball can roll, as indicated in the figure(Figure 1). If a ball of mass 2M and radius r=R/4 is released as shown, the system rocks back and forth until it comes to rest with the ball at the bottom of the hoop. Part A. When the ball comes to rest, what is the x coordinate of its center? Express your answer in terms of the variables R and M.
Part A: There is only one external force, the gravitational force which exerts only in vertical direction.Center of mass is in the same horizontal position before and after the ball is released, though it will move vertically, it is stationary on the x-axis. We know ha after settling sown hoop and the ball will have the same x- coordinate Therefore initial Centre of Mass = final Centre of Mass [ (0R)(M) + (0.75R)2M] / [M +2M] = (Mx + 2Mx)/ [M + 2M] 1.5 R = 3x x = 0.5*R or (1/2)*R
The human head can be considered as a 3.3-kg cranium protecting a 1.5-kg brain, with a small amount of cerebrospinal fluid that allows the brain to move a little bit inside the cranium. Suppose a cranium at rest is subjected to a force of 2600 N for 7.5 ms in the forward direction. Part A. What is the final speed of the cranium? Part B. The back of the cranium then collides with the back of the brain, which is still at rest, and the two move together. What is their final speed Part C. The cranium now hits an external object and suddenly comes to rest, but the brain continues to move forward. If the front of the brain interacts with the front of the cranium over a period of 13 ms before coming to rest, what average force is exerted on the brain by the cranium?
Part A: using the impulse and momentum of the system equation. FdeltaT=mv (2600)(7.5x10^-3)=3.3 v= (2600)(7.5x10^-3)/3.3 = 5.9 m/s Part B: Apply the conservation of momentum m1V=(m1+m2)V' (3.3)(5.9)/(1.5+3.3) V'= 4.1 m/s Part C: The acceleration during the proces a= vf-vi/t = 0-4.1/13x10^-3 = -315.4 Force= F=ma = (1.5)(-315.4) = -470 N
The two masses in the Atwood's machine shown in the figure are initially at rest at the same height. After they are released, the large mass, m2, falls through a height h and hits the floor, and the small mass, m2, rises through a height h. Part A: Find the speed of the masses just before m2 lands, giving your answer in terms of m1, m2, g, and h. Assume the ropes and pulley have negligible mass and that friction can be ignored. Part B: Evaluate your answer to part A for the case h = 1.7 m , m1 = 3.5 kg , and m2 = 4.3 kg.
Part A: vf = 2gh(m2−m1m2+m1) {all under the square root function} Part B: vf= 1.8 m/s
A block of weight w = 40.0 N sits on a frictionless inclined plane, which makes an angle θ = 21.0 ∘ with respect to the horizontal, as shown in the figure. (Figure 1)A force of magnitude F = 14.3 N , applied parallel to the incline, is just sufficient to pull the block up the plane at constant speed. Part A: The block moves up an incline with constant speed. What is the total work Wtotal done on the block by all forces as the block moves a distance L = 4.20 m up the incline? Include only the work done after the block has started moving at constant speed, not the work needed to start the block moving from rest. Part B: What is Wg, the work done on the block by the force of gravity w as the block moves a distance L = 4.20 m up the incline? Part C: What is WF, the work done on the block by the applied force F as the block moves a distance L = 4.20 m up the incline? Part D: What is WN, the work done on the block by the normal force as the block moves a distance L = 4.20 m up the inclined plane?
Part A:Since the block is moving at a constant speed there is no change in kinetic energy. Since work is the change in kinetic energy and there is no change, the network on the block is zero. Part B: Wg=-mgh= -wlsintheta Part C: Wf= FL=FLcostheta Part D: The normal force is perpendicular to the direction of motion, therefore it does no work on the block.
Potential Energy
Potential energy is a concept that builds on the work-energy theorem, enlarging the concept of energy in the most physically useful way. The key aspect that allows for potential energy is the existence of conservative forces, forces for which the work done on an object does not depend on the path of the object, only the initial and final positions of the object. The gravitational force is conservative; the frictional force is not. The change in potential energy is the negative of the work done by conservative forces. Hence considering the initial and final potential energies is equivalent to calculating the work done by the conservative forces. When potential energy is used, it replaces the work done by the associated conservative force. Then only the work due to nonconservative forces needs to be calculated. In summary, when using the concept of potential energy, only nonconservative forces contribute to the work, which now changes the total energy: Kf+Uf=Ef=Wnc+Ei=Wnc+Ki+Ui,where Uf and Ui are the final and initial potential energies, and Wnc is the work due only to nonconservative forces. Now, we will revisit the falling stone example using the concept of potential energy.
