# Physics Exam 2 F2020

## Unlock all answers in this set

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Using Vectors on Motion Diagrams
â€˘In two dimensions, an object's displacement is a vector: â€˘The velocity vector is simply the displacement vector multiplied by the scalar (1/change in t) â€˘Consequently the velocity vector points in the direction of the displacement. velocity vector= d vector (both x and y components)/ change in time
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the velocity vector points in the direction of the
displacement
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Acceleration in Two Dimensions
â€˘The vector definition of acceleration is a straightforward extension of the one-dimensional version: change in acceleration vector=change in velocity vector (vf-vi)/ change in time â€˘There is an acceleration whenever there is a change in velocity. Velocity can change in either or both of two possible ways: 1. The magnitude can change, indicating a change in speed. 2. The direction of motion can change, indicating a change in direction.
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â€˘There is an acceleration whenever there is a change in velocity. Velocity can change in either or both of two possible ways:
â€˘There is an acceleration whenever there is a change in velocity. Velocity can change in either or both of two possible ways: 1. The magnitude can change, indicating a change in speed. 2. The direction of motion can change, indicating a change in direction.
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A particle undergoes acceleration a ? while moving from point 1 to point 2. Which of the choices shows the velocity vector (v_2 ) ? as the object moves away from point 2
find Vf component (y component) of acceleration vector
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The diagram shows three points of a motion diagram. The particle changes direction with no change of speed. What is the acceleration at point 2?
need the acceleration vector which makes the two velocity vectors equal (both the V1 from 1-2 and the Vf that is drawn in)
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Reminder of Motion Diagram for Circular Motion
acceleration is perpendicular to velocity -velocity is tangent to the circle and a is perpendicular to v, pointing toward the center to find the change in velocity vector, subtract the velocity at point 1 from the velocity at point2 -the acceleration vector points in the same direction as the change in velocity vector, so toward the center of the circel
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period-T
time it takes object to complete a full circle [T]=S
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frequency-f
the number of revolutions per time period 1/s= hertz Hz
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centripetal acceleration
acceleration toward the center of a curved or circular path m/s^2 b/c is V^2/r which is (m^2/s^2)/m
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circumference
2?r or 2d
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A car is turning a tight corner at a constant speed. A top view of the motion is shown in the figure. The velocity vector for the car points to the east at the instant shown. The direction of the acceleration is a.N b.E c.S d.W
south
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A runner is running at a uniform pace on a circular track of radius 50 m. It takes her 314 seconds to complete one lap. Determine her speed.
r=50m t=314sec velocity=distance travelled/time acceleration is constant c=2pir =2pi(50)=100pi =100pi/314 sec =1m/s
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force
Forces are pushes or pulls. Every force acts on an object. Every force has an agentâ€”something that does the pushing or pulling.
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(just one for us) Long-range or non-contact forces
gravitational (weight of object)
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contact forces
These exist only as the objects are touching each other. v actual touchingâ€”from a shove or a kick v friction v tension in a rope or chain v forces from floors, tables, or walls v drag forces such as air resistance thrust from rocket engines NO Force is objects are NOT touching--it's the interaction that is the force
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Newton's First Law of Motion Law of Inertia
If the net force acting on an object is zero. â€˘ and if this object is at rest, it will remain at rest. â€˘ and if this object is in motion, it will continue to move in a straight line at a constant speed.
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thought experiment: somebody has a hockey stick and a hockey puck-can give same amount of velocity to object when they hit it -in the grass, hit hockey puck -on rough ice -smooth ice -frictionless ice
stops b/c of friction -goes farther, but still slows to a stop -on smooth goes farther -on frictionless, speed never changes and once hickey stick isn't touching it, the net force is zero and the speed is maintained-would go forever
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HP nightbus movie example
Harry standing with net force of zero, bus moves forward and he slams into back b/c not enough friction -Harry then moves at same constant speed as bus -bus slams on brakes, HP moves forward and slams into front
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Net force (a vector)
vFind the vector sum of all forces exerted on an object. ? not interested in forces exerted by the object. vThe vector sum may be expressed in component form or as a magnitude and direction.
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The figure shows the top view of a box, in which only the forces shown are acting on the box. Determine the net force in component form. 5N to left and 10N to R
5N to right
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The figure shows the top view of a box, in which only the forces shown are acting on the box. Determine the net force in component form. 4N up, 6N left, 8N down
6N left and 4N down in component form
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Newton's Second Law, preliminary
If the net force on an object is not zero, the velocity of the object changes. -change in velocity per unit of time is acceleration -acceleration has the same direction as the net force
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acceleration has the same direction as the
net force
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The figure shows the top view of a box, in which only the forces shown are acting on the box. Determine the net force in component form. This box is moving to the right. Which of the motion diagrams agrees with the motion of the block. 5N to left and 10N to right
net force in component form direction of net force gives direction of acceleration net force is 5N to the right, motion diagram shows speeding up
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The figure shows the top view of a box, in which only the forces shown are acting on the box. Determine the net force in component form. This box is moving to the left. Which of the motion diagrams agrees with the motion of the block. 5N to left and 10N to right
but box is moving to L -negative velocity (because moving to L) -acceleration is 5n to the R, so opposite signs, so slowing down
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does the motion of the object have to be in the direction of the net force?
no, could be, but not always
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Weight
â€˘The gravitational pull of the earth on an object on or near the surface of the earth is called weight. â€˘The agent for the weight forces is the entire earth pulling on an object. â€˘An object's weight vector always points vertically downward, no matter how the object is moving.
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Spring Force
â€˘Springs come in in many forms. When deflected, they push or pull with a spring force. object=box agent=spring
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Tension Force
â€˘When a string or rope or wire pulls on an object, it exerts a contact force that we call the tension force. The direction of the tension force is always in the direction of the string or rope agent=rope acting on the sled -rope exerts a tension force on the sled
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Normal Force
â€˘The force exerted on an object that is pressing against a surface is in a direction perpendicular to the surface. â€˘The normal force is the force exerted by a surface (the agent) against an object that is pressing against the surface. agent: desk object: book
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Friction can occur when object is ____ or _____
moving or sedentary
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if sled is moving to the right but slowing down, V vector is directed toward the right, but fk vector is to the ___ b/c a kinetic friction force
left b/c kinetic friction force is slowing the sled down
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woman pulling to the left, but the crate does not move b/c
a static friction force directed to the right opposes that motion
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Drag
â€˘The resistive force of a fluid (like air or water) on a moving object is called drag. â€˘Like kinetic friction, drag points opposite the direction of motion. You can neglect air resistance in all problems unless a problem explicitly asks you to include it
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Thrust
â€˘Thrust is a force that occurs when a jet or rocket engine expels gas molecules at high speed. (octopus or squid also) â€˘Thrust is a force opposite the direction in which the exhaust gas is expelled. object=rocket agent=molecules in rocket fuel as they come out
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A red-tailed hawk that weighs 8 N is gliding at constant speed. What is the net force acting on the hawk? a.8 N in the downward direction b.8 N in the direction the hawk is flying c.8 N in the upward direction d.0 N Draw the motion diagram for the hawk as it is flying with constant speed.
net force is zero on the hawk, because speed is constant dots are equally spaced
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Emily pushes on a box with a horizontal force of 10 N. The box does not move. What is the net force on the box? a.10 N in the direction of the push b.0 N c.10 N opposite the direction of the push d.10 N down
net force is 0 N because static friction in 10N in opposite direction of push
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constructing a free body diagram A bungee jumper has jumped off of a bridge and is nearing the bottom of her fall. What forces are exerted on the bungee jumper?
-taking/freeing from the context of the problem and representing environment with forces object is bungee jumper tension from rope cuts through circlr
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steps of free body diagram
1. draw boundary around man (what we're studying) pin point the object and its environment 2. ID what forces: name and label--tension from rope, earth on man 3. name and label long range force (not physical)--gravity
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Ron pushes to the right on a box. The box does not move. Draw a free body diagram of the box. Identify the object being acted on AND the agent of the force.
agent is ron object is box
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a. agent: normal force of hand on object: rock equal vector length of both Normal and Weight b/c net force is zero b. weight force is the same, but N is longer c. no N force, b/c no contact with the hand; just W d. only thing acting on it is weight force
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An elevator, lifted by a cable, is moving upward and slowing. â€˘What is the direction of the acceleration? â€˘What is the direction of the net force? â€˘Which is the correct free-body diagram? Be prepared to identify the agent of any forces shown in your selected diagram.
-negative negative, down (direction of acceleration has to be the direction of the net force) T rope up and W down which is longer
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A ball, hanging from the ceiling by a string, is pulled back and released. We are interested in the time just after the ball is released. â€˘What is the direction of the acceleration? â€˘What is the direction of the net force? â€˘Which is the correct free-body diagram? Be prepared to identify the agent of any forces shown in your selected diagram.
positive pos
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A car is parked on a hill. â€˘What is the direction of the acceleration? â€˘What is the direction of the net force? â€˘Which is the correct free-body diagram? Be prepared to identify the agent of any forces shown in your selected diagram.
tilted axis -force from car touching incline -acceleration is zero b/c not moving -net force is thus zero -friction force points uphill b/c if started, would roll down, so friction force is opposite; normal force points up, W straight down
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A car is towed to the right at constant speed. â€˘What is the direction of the acceleration? â€˘What is the direction of the net force? â€˘Which is the correct free-body diagram? Be prepared to identify the agent of any forces shown in your selected diagram.
zero zero D-all forces are equal, b/c net force is zero
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newton's second law
â€˘A force causes an object to accelerate. â€˘The acceleration a is directly proportional to the force F and inversely proportional to the mass m: acceleration=fnet/mass â€˘The direction of the acceleration is the same as the direction of the force:
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If a golf ball and bowling ball have same hit/Fnet, the ____________________ ball has larger acceleration.
golf ball has larger acceleration b/c smaller mass
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what is a Newton?
unit of force [F-fnet]=kg x m/s^2 = N
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At the instant that these two forces are applied, the 10 kg box has a velocity of zero. What is its velocity 5 seconds later? 5N to left and 10 N to right
net force is 5N to the right mass is 10kg acceleration is 5N/10kg = 0.5 m/s^2 now back to constant acceleration to find velocities change in V = (0.5 m/s^2)(5s) = + 2.5 m/s at 5 seconds
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t the instant that these two forces are applied, the 10 kg box has a velocity of -4 m/s. What is it's velocity 5 seconds later? 5N to left and 10N to right
acceleration is 0.5 m/s^2 velocity calculated before as +2.5m/s change in V = Vf-Vi Vf- (-4)=2.5m/s
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A 5-kg block is traveling such that v_x=+4 m/s and v_y=+4 m/s. Two forces are then applied as shown in the diagram. â€˘What are the x- and y- components of the acceleration? What are the x- and y- components of the velocity in 5 seconds? 10N up and 20N to the right
accel in y: 10N/5kg= 2 m/s^2 accel in x: 20/5= 4 velocity: in y: 2 m/s^2 (5s)=change in velocity=10m/s started at v=4 vf in y=14m/s in x: change in velocity=(4m/s^2)(5s)= 20 m/s vf= 20 + 4 = 24m/s
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A 10-kg box is traveling to the right with a speed of 30 m/s. â€˘What are the x- and y- components of the acceleration? â€˘ â€˘What are the x- and y- components of the velocity after 2 seconds? force 100 N down
acceleration in y: -100N/10kg = -10m/s acceleration in x=0 (constant speed) velocity after 2 seconds in y: change in velocity= acceleration x change in time =(-10m/s^2)(2s)= -20m/s Vf-0=20m/s in x: chnage in V= 0x2=0 Vf-30=0 vf=30m/s
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Scallops eject water from their shells to provide a thrust force. The graphs shows a smoothed graph of actual data for the initial motion of a 25 g scallop speeding up to escape a predator. What is the magnitude of the net force needed to achieve this motion? How does this compare to the 0.25 N weight of the scallop?
given velocity -find a to find net force scallop weighs 25 g--must be in KG--0.025kg acceleration= (0.2-0)/(0.2-0) = 1m/s^2 a= fnet/m 1=fnet/0.025kg = 0.025N how does this compare to the 0.25 N weight of scallop? RATIO: divide 0.025N/0.25N = 0.1 how big is the force relative to the weight force
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Newton's Third Law
â€˘Motion often involves two or more objects interacting with each other. â€˘As the hammer hits the nail, the nail pushes back on the hammer. â€˘A bat and a ball, your foot and a soccer ball, and the earth-moon system are other examples of interacting objects.
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Interacting Objects: interaction and action/reaction pair
â€˘An interaction is the mutual influence of two objects on each other. â€˘The pair of forces shown in the figure is called an action/reaction pair.
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â€˘An action/reaction pair of forces exists as a ___, or ___ at all.
pair or not at all
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Â§The two members of an action/reaction pair act on ___ _____ objects. Â§The two members of an action/reaction pair point in ____ directions and are ____ in magnitude
Newton's third law Every force occurs as one member of an action/reaction pair of forces. Â§The two members of an action/reaction pair act on two different objects. Â§The two members of an action/reaction pair point in opposite directions and are equal in magnitude.
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A bat hits a ball. Imagine you are viewing this from the end of the bat. â€˘Draw a free body diagram of the bat. â€˘ â€˘Draw a free body diagram of the ball. â€˘ â€˘Identify third law pairs. â€˘ What is the third law pair for the weight of the bat?For the ball?
FB diagram of bat: Fball on bat to left nad W down FB of ball: F bat on ball to right and W down F bat on earth up from earth and F ball on earth up from earth 3rd law pairs: Fball on bat and Fbat on ball -Fball on earth Fbat on earth person holding bat: F person on bat and third law pair is Fbat on person
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A mosquito runs head-on into a truck. Splat! Which is true during the collision? A. The mosquito exerts more force on the truck than the truck exerts on the mosquito. B. The truck exerts more force on the mosquito than the mosquito exerts on the truck. C. The mosquito exerts the same force on the truck as the truck exerts on the mosquito. D. The truck exerts a force on the mosquito but the mosquito does not exert a force on the truck. E. The mosquito exerts a force on the truck but the truck does not exert a force on the mosquito.
C. The mosquito exerts the same force on the truck as the truck exerts on the mosquito.
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mass vs weight
Mass is an inherent property of an object. The object can be relocated and the mass will not change. Weight is the force on the object due to the gravitational attraction of a large astronomical body. That body can be the Earth, the moon, Mars, etc. Weight is not an inherent property of the object. Mass and weight are linked through w=mg. Here on earth, we sometimes interchange the words in our everyday conversation.
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Mass and weight are linked through
Mass and weight are linked through w=mg
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1 lb in N
4.45 N
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110lb person in N 1lb object in N
500N 5N
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Newton's Universal Law of Gravitation
Mass attracts mass. F(1 on 2)= F(2 on 1)=G (m_1 m_2)/r^2 Where G is the universal gravitation constant G=6.67 x 10^(-11) (N m^2)/?kg?^2 . m1 and m2 are masses. r is the distance between the masses.
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what is Fearth on book?
Fe on b= 6.67 x 10^-11 Nm^2/kg^2 ((5.98x10^24 x 1.0kg)/(6.37 x 10^6 m)^2) = 9.8N (acceleration due to gravity comes from this
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When you are near an astronomical body: F=m (G m_(ast body)/r^2 )
And we call the force, weight. W=mg
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when Fnet = 0 Applying Newton's Laws
constnat speed or stationary if fnet=0 total Fx=0 total Fy=0 equilibrium when net force=0; static or dynamic if net force=0, all forces in the x and y direction = 0
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when Fnet does not equal zero Applying Newton's Laws
accelerating total Fx=m(ay) total Fy=m (ax) NOT at equilibrium
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A car with a mass of 1500 kg is being towed at a steady speed by a rope held at a 20Â° angle from the horizontal. A friction force of 320 N opposes the car's motion. What is the tension in the rope?
2 methods
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â€˘At the first plateau, the jumper is standing still. Draw a FBD for the jumper and determine the jumper's weight and his mass. â€˘ â€˘ â€˘At the second plateau, the jumper is accelerating upwards. Draw an FBD and determine the net force acting on the jumper. Hint: It is not 2000 N. â€˘ â€˘Determine the jumper's acceleration. â€˘ â€˘Determine the jumper's speed as he leaves the ground. Let's assume that he accelerates for 0.25 s. â€˘ â€˘Determine the jumper's maximum height.
-at first plateau, standing still-FBD N and W n=weight=1000N weight=mg 1000N=m(10N/kg) m-100kg at second plateau: N is largeer than weight (2000N vs. 1000N) Fnet=2000-1000=1000N up acceleration: 1000N=a(100kg) a=10m/s^2 accelerates for 0.25sec: speed/velocity as leaves ground: change in V=(10m/s^2)(0.25s)= 2.5m/s starts from 0, so Vf=2.5m/s max height: at top of jump Vf=0 vi=2.5m/s 0^2 - 2.5^2 = 2 (-9.8m/s^2)(change in y) change in y= 0.319m
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in order to jump, must push off floor if you push off floor then the floor pushes back
to get N force bigger than W force for acceleration in upward direction
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Example 5.8 Anjay's mass is 70 kg. He is standing on a scale in an elevator and is moving at 5.0 m/s. As the elevator slows to a stop, the scale reads 750 N. Before it stopped, was the elevator moving up or down? Hint: Draw a FBD. The scale reading is the normal force exerted by the floor of the elevator on Anjay.
nscale on A is 750N WEonA is 700N was going up, but elevator slowed down to a stop, so a is positive but velocity is neg apply newton's second law
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astronaut bones in space
do not have the weight force compressing the bone
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Apparent Weight
Your sensation of weight is due to the contact forces supporting you. In most situations we call this supporting force the normal force. However, when we are concentrating on a human and what they are "feeling", we frequently refer to this as the apparent weight. To find the apparent weight of somebody in an elevator, draw a FBD and then apply Newton's second law.
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Normal force for Objects on Inclines
ignore friction rotate picture normal points up along y axis and w points out at an angle components: WII to interface is y component Wperp to interface is x componenet acceleration perpendicular to the interface has to equal zero n=mgcos(theta) if on incline, normal force is smaller than on flat ground cos(theta < 1
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5.6 tension on a rope while towing a car: it is accelerating car with mass= 1500kg towed by a rope at 20 degree angle from the horizontal friction force is 320N opposes car's motion what is th etnesion in the rope of the car goes from rest to 12m/s in 10 sec
acceleration is (12m/s)(10s)= 1.2 m/s^2 break T into components Ty=Tsin20 Tx=Tcos20 x direction: accelerating so Tx>320N in y direction: not accelerating=0 ax=1.2m/s^2 total Fx=m(ax) m(ax)=Tcos(20)-320 (1500kg)(1.2m/s^2)=Tcos(20)-320 T=2256.06N
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A box is being pulled to the right at steady speed by a rope that angles upward. In this situation: A. n > mg B. n = mg C. n < mg D. n = 0 E. Not enough information to judge the size of the normal force
C. n < mg b/c n+T=mg (the weight)
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The box is sitting on the floor of an elevator. The elevator is accelerating upward. The magnitude of the normal force on the box is
A. n > mg normal force comes from something physically underneath the object
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An 800-N bicyclist is accelerating up a hill. In this situation w= 800N incline is 20 degrees
n=mgcos(theta) cos(theta) cannot be >1 on flat surface, cos(0)=1, so n=mg n
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static friction If the box is stationary, then ???F ?=0?. There must be a second horizontal force exerted on the box so that ???F ?=0?. What is that force? What is its magnitude and what is its direction?
Friction that acts on objects that are not moving A force is applied to the box, but the box remains stationary. If the box is stationary, then ???F ?=0?. There must be a second horizontal force exerted on the box so that ???F ?=0?. What is that force? What is its magnitude and what is its direction? Fnet is 0
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maximum static friction force
the maximum friction force measured just before the impending motion of an object A system can only supply a certain amount of static friction. This is called the maximum static friction, fs,max. If the magnitude of the force F is larger than the magnitude of the maximum static friction for the system, the box will begin to slide. The maximum static friction amount for a system can only be determined by experimental investigation. (empirically). f_(s,max)= ?_s n
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f_(s,max)= ?_s n Âµs n
Âµs is called the coefficient of friction and is found in a table. n is the magnitude of the normal force. Remember it may or may not be equal to mg.
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kinetic friction
Friction between moving surfaces Kinetic friction arises when two surfaces are sliding past each other. Kinetic friction depends on the materials of the two surfaces and must be determined empirically (meaning--in the lab). f_k= ?_k n
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f_k= ?_k n
Âµk is called the coefficient of kinetic friction. It must be looked up in a table. n is the magnitude of the normal force.Remember it may or may not be equal to mg
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static friction constant is always ____ than uk
bigger at max static friction, starts to slide and don't have to push as much to keep moving b/c uk
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how friction develops
These two surfaces are sliding or trying to slide past each other. The atomic sized protrusions and valley interact with each other. very few points are actually in contact b/c atomically there are bumps
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Carol wants to move her 32 kg sofa to a different room in the house. She places "sofa sliders," slippery disks with ?k = 0.08 on the carpet, under the feet of the sofa. She then pushes the sofa at a steady 0.40 m/s across the floor. How much force does she apply to the sofa?
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A car with a mass of 1500 kg is being towed at a steady speed by a rope held at a 20Â° angle from the horizontal. The coefficients of friction are listed below. The tension in ?_s=0.8 ?_k=0.6 ?_r=0.3 Â§Draw FBD Â§Determine which coefficient of friction applies. Â§Write your ?F_x=m a_x Â§Write your ?F_y=0
use rolling friction because being pulled static friction if driven
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A student pulls with a 100-N horizontal force on a rope attached to a wall. The tension in the rope is (a)Less than 100 N. (b)Equal to 100 N. Greater than 100 N.
equal to 100N -must cut rope to see internal tension force
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Two students are pulling on a rope. Each pulls with a 100 N force. The tension in the rope is (a)Less than 100 N. (b)Equal to 100 N. Greater than 100 N.
equal to 100N
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A rope connects two boxes (1 10kg, 1 20kg). A second rope is attached to the yellow box and maintains a tension of 300 N. The floor is frictionless. Which of the following statements is true? (There may be more than one true statement.) a.The acceleration of both blocks is the same. b.The velocity of both blocks is the same. c.The acceleration of the 20-kg-block is less than that of the 10-kg-block. d.The velocity of the 20-kg-block is less than that of the 10-kg-block. e.The acceleration of the 20-kg-block is more than that of the 10-kg-block. f.The velocity of the 20-kg-block is more than that of the 10-kg-block.
a.The acceleration of both blocks is the same. b.The velocity of both blocks is the same. rope does not crumple up and is not elastic
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A rope connects two boxes(1 10kg, 1 20kg). A second rope is attached to the yellow box and maintains a tension of 300 N. The floor is frictionless. (a)Determine the acceleration of each block. Determine the tension in the rope attaching the green box to the yellow box.
acceleration is the same: so circle both boxes, draw FBD with n, t=300N, and w accerleration = 300N/60kg= 5m/s^2 determine tension in rope: cute off one box: now n, T, and W ax=fnetx/m=5m/s^2 5=T/20kg T=100N
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Conservation Laws Conservation of Linear Momentum Conservation of Angular Momentum Conservation of Energy Conservation of Mass what does conservation mean? what is the system?
The word conservation means that the quantity does not change. What we usually mean is that under some set of conditions, the quantity does not change. Systemâ€”composed of interacting objects
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Collisions
short-duration interaction between two objects The interaction between the foot and the soccer ball is a collision.
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impusle
Impulse, J, is the area under the curve
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impulse is also
The F vs t curve is complicated, so we use the concept of average force. = Fave ?t
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impulse causes
Impulse causes a change in the motion of an object.
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The figure shows a force versus time data for the bounce of a ball. â€˘What is the impulse on (or delivered to) the ball? â€˘ â€˘What is the average force on the ball?
J=area under curve =.5(0.008s)(300N) =1.2N average force: Fave x change in time=J 1.2N/0.008s=Fave Fave=150N
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when the stick is in contact with the puck, it exerts a force Fave on the puck, causing it to accelerate
Fave only acts for a short time. It is an impulsive force. We want to find out how it changes the velocity of the puck. We will start analyzing the problem with Newton's second law. As a result of the analysis, we will find a new quantity called momentum.
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momentum Impulse and momentum are vector quantities.
initial momentum + impulse= final momentum change in momentum=impulse change in momentum=Fave x change in t
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Rank the objects according to the magnitudes of the momentum, smallest to largest.
mass x velocity= momentum
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Two 1.0 kg stationary cue balls are struck by cue sticks. The cues exert the forces shown. Which ball has the greater final speed?
same final speed-b/c impulse is the same b/c area is same
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A 2.0 kg object moving to the right with speed 0.50 m/s experiences the force shown. What are the object's speed and direction after the force ends?