lab ch 3 – Flashcards
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Vector quantity
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quantity having both magnitude and direction
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Displacement (Δx)
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Change in position, vector quantity
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Resultant Vector
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What is Vector R?
Max = Vector A + Vector B
Min = A - B
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Velocity (instant... then normal below)
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'Displacement over time, vector quantity'
v = Δx / Δt
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Average speed
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Total distance / Total time
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Acceleration
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'Change in velocity over time, vector quantity'
instant is above... normal:
Δv / Δt
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0, 30, 15, 15, 0, 50, 0, 0, (-9.81)
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xᵩ=?, x=?, vᵩ=?, v=?, a=?, Yᵩ=?, Y=?, Vᵩ=?, A=?
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θ
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angle in projectile motion
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0, 20, 20Cos(45), 20Cos(45), 0, 60, 0, 20Sin(45), (-9.81)
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xᵩ=?, x=?, vᵩ=?, v=?, a=?, Yᵩ=?, Y=?, Vᵩ=?, A=?
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A(x) = ACosθ, A(y) = ASinθ
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The above is a right Δ. solve for A(x) and A(y) *Note: refer to the angle at the vertex of A and A(x)
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Work on a Block Sliding Up a Frictionless Incline:
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Block weight 15.0 N sits on a frictionless inclined plane, which makes an angle θ = 23.0° with respect to the horizontal, as shown in the figure. A force of magnitude F = 5.86 N, applied parallel to the incline, is just sufficient to pull the block up the plane at constant speed.
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a) The block moves up an incline with constant speed. What is the total work W. Total done on the block by all forces as the block moves a distance L = 3.40 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.
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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 net work on the block is zero.
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b) What is Wg, the work done on the block by the force of gravity w‖ as the block moves a distance L = 4.50m up the incline?
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1) Find the component of the gravitational force parallel to the plane
What is w||, the magnitude of the component of the force of gravity along the inclined plane? →w parallel to the inclined plane has magnitude given by w‖=wsinθ. = 40*sin(22) = 15.
2) Wg = -mgy = -15*4.5 = -67J
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What is Wƒ, the work done on the block by the applied force F→ as the block moves a distance = 4.5m up the incline?
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Since we found the net work to be zero (in Part A), the work done by the applied force has to offset the work done by gravity: Wg=−Wƒ. ∴ 67J
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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
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Ki+Ui=Kf+Uf,
K=½mv²
Potential: Ug = mgh
Elastic potential energy: For a spring with a force constant k, stretched or compressed a distance x, the associated elastic potential energy is
Ue=½kx²
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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?
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Think about these questions:
Are there any nonconservative forces acting on the block during this part of the trip?
Are there any objects involved that can store elastic potential energy?
Is the block changing its height?
Is the block changing its speed?
½mv₁² + mgh₁ = ½mv₂² + mgh₂
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c) 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
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½mv₁² + mgh₁ + ½kx₁² + Wnc = ½mv₂² + mgh₂ + ½kx₂²
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.
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d) Using conservation of energy, find the speed vb of the block at the bottom of the ramp.
Express your answer in terms of some or all the variables m, v, h and and any appropriate constants.
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PE₀ + KE₀ = PEƒ + KEƒ
At the top of the ramp, potential energy will be at a maximum and kinetic energy will be at a minimum. At the bottom of the ramp, kinetic energy will be at a maximum and potential energy will be minimum. But although the type of energy changes, the total amount of energy remains the same:
mgh + ½mv² = mgh(b) + ½mv(b)²
Since the height at the bottom is zero, we can eliminate potential energy from the right side of the formula:
mgh + ½mv² = ½mv(b)²
gh + ½v² = ½v(b)²
2gh + v² = v(b)²
v(b) = √(v² + 2gh)
== √(v² + 2gh)
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As the block slides across the floor, what happens to its kinetic energy K, potential energy U, and total mechanical energy E?
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K decreases
U stays the same
E decreases
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g) What force is responsible for the decrease in the MECHANICAL ENERGY of the block?
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friction
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⊕ Mechanical Εnergy →
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#####NEXT#####
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Mechanical energy of a system
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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
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A block of mass m slides at a speed v along a horizontal smooth table. It next slides down a smooth ramp, descending a height h, and then slides along a horizontal rough floor, stopping eventually. Assume that the block slides slowly enough so that it does not lose contact with the supporting surfaces (table, ramp, or floor).
You will analyze the motion of the block at different moments using the law of conservation of energy.
a) Which word in the statement of this problem allows you to assume that the table is frictionless?
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Although there are no truly "frictionless" surfaces, sometimes friction is small enough to be neglected. The word "smooth" often describes such low-friction surfaces. Can you deduce what the word "rough" means?
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h) Find the amount of energy E (MECH ENEG) dissipated by friction by the time the block stops. Express your answer in terms of some or all the variables m, v, and h and any appropriate constants.
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Friction will slow the block to zero, so it will dissipate all of the kinetic and potential energy that the block has at the top of the ramp.
= ½mv² + mgh
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⊕ WORK Ƒriction
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###NEXT###
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5.33: Find the work done by friction. (SI: J)
A 1.37kg book slides 1.26m along a level surface. The coefficient of kinetic friction between book and surface is 0.154.
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work done = force x distance
mg = 1.37kg x 9.8 = 13.426N
13.426 x 0.154 = force required to overcome friction = 2.07N (rounded)
(0 - 2.07) = -2.07 N (force of friction)
-2.07 x 1.26
≅ -2.61J (answer)
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⊕ Spring Constant / Equilibrium
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###NEXT####
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Spider Silk: Spider silk is one of the most remarkable elastic materials known. Consider a silk strand suspended vertically with a 0.35-g fly stuck on the end. With the fly attached, the silk measures 28.0 cm in length. The resident spider, of mass 0.66 g, senses the fly and climbs down the silk to investigate. With both spider and fly at the bottom, the silk measures 37.5 cm
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So a mass of 0.66g caused a change in length of (37.5 - 28.0) = 9.5 cm
(9.5 ÷ 100) = 0.095 m [to match ƒg units]
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a) Find the spring constant of the silk.
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So a mass of 0.66g caused length Δ of (37.5 - 28.0) = 9.5 cm ↔ 0.095m {gravity is in meters}
Recall that the spring constant is normally expressed in Newtons per meter
(ƒg =9.8 and 0.66 g = 0.00066 kg )
0.00066 * 9.8 = 0.006468 N,
Spring constant = (0.006468 ÷ 0.095 Nm⁻¹) = 0.068 Nm⁻¹
≡ 6.8 * 10⁻² (N/m)
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b) Find the equilibrium length of the silk.
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Fly's weight in N = 0.00035 * 9.8
= 0.00343 N
∴ amount the fly stretches the thread must be: Mass(N) ÷ Spring Constant
= 0.00343 ÷ 0.068 = 0.050 m
= 5 cm
Equilibrium length = 28 cm- 5cm
≅ 23 cm;
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Force - Push/pull
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One of the biggest mistakes you may make is to think of a force as "something an object has." In fact, at least two objects are always required for a force to exist.
Each force has a direction: Forces are vectors. The main result of such interactions is that the objects involved change their velocities: Forces cause acceleration. However, in this problem, we will not concern ourselves with acceleration--not yet.
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Which object exerts a downward force on the book?
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earth
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What is the downward force acting on the book called?
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weight
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Which object exerts an upward force on the book?
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surface of the table
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The upward force acting on the book is ___ force
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contract
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_____ is a pulling force.
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Tension
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____ is a pushing force.
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normal [force]
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⊗NEXT
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----->
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A string is attached to a heavy block. The string is used to pull the block to the right along a rough horizontal table.....
Which object exerts a force on the block that is directed toward the right?
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string
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The force acting on the block and directed to the right is____
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contract-forrce..
To exert a tension force, the object must be connected to (i.e., touching) the the object it is acting upon..
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What is the force acting on the block and directed to the right called?
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tension
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Which object exerts a force on the block that is directed toward the left?
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the surface of the table
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The force acting on the block and directed to the left is a ___ force.
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contract
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What is the force acting on the block and directed to the left called?
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friction
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Now consider a slightly different situation. The same block is placed on the same rough table. However, this time, the string is disconnected and the block is given a quick push to the right. The block slides to the right and eventually stops. The following questions refer to the motion of the block after it is pushed but before it stops.
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Hint 1: Include only those forces after the block is pushed. Try listing all of the forces and carefully thinking about what object causes them. Recall that an object cannot exert a force on itself or carry a force around from an earlier time.
Hint 2: Remember to include only horizontal forces. Try listing all of the forces and carefully thinking about what object causes them. Recall that an object cannot exert a force on itself or carry a force around from an earlier time.
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How many forces are acting on the block in the horizontal direction?
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1 -----
Once the push has ended, there is no force acting to the right: The block is moving to the right because it was given a velocity in this direction by some force that is no longer applied to the block (probably, the normal force exerted by a student's hand or some spring launcher).
Once the contact with the launching object has been lost, the only horizontal force acting on the block is directed to the left--which is why the block eventually stops.
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Newton's 1st Law
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Newton's Principia states this first law of motion:
An object subject to no net force maintains its state of motion, either at rest or at constant speed in a right line.
This law may be restated as follows: If the sum of all forces acting on an object is zero, then the acceleration of that object is zero. Mathematically this is just a special case of the 2nd law of motion, "F=ma" when F=0.
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Newton's 1st Law -> two cases
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1) If the FORCE (i.e., sum of all forces) acting on an object is ZERO, the object will KEEP MOVING with constant velocity (which may be zero).
2) If an object is moving with CONSTANT VELOCITY, that is, with ZERO ACCELERATION, then the net force acting on that object MUST BE ZERO.
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If a car is moving to the left with constant velocity, one can conclude that ______
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e net force applied to the car is zero.
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An object cannot remain at rest unless ____
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the net force acting on it is zero.
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An object will have constant acceleration if ____
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the net force acting on it is constant in magnitude and direction.
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⊗NEXT ---->
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3rd Law
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Newton's 3rd Law Discussed
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To understand Newton's 3rd law, which states that a physical interaction always generates a pair of forces on the two interacting objects.
In Principia, Newton wrote:
To every action there is always opposed an equal reaction: or, the mutual actions of two bodies upon each other are always equal, and directed to contrary parts.
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3rd Law Def
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Whatever the physical cause of the interaction, the force acting on object A due to object B is EQUAL in MAGNITUDE and OPP in DIRECTION to the force acting on object B due to object A.
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Every force has one and only one 3rd law pair force. [T/f]?
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true
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The two forces in each pair can act on the same object or on different objects. [T/f]?
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false
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The two forces in each pair may have different physical origins (for instance, one of the forces could be due to gravity, and its pair force could be a normal contact force).
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FALSE
Each type of physical interaction (e.g., gravity, friction, or electrostatic) always produces a pair of forces. If it did not, the center of mass of the universe would accelerate as a result of the unmatched forces.
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The two forces of a 3rd law pair always act on different objects.
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true
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Given that two objects interact via some force, the accelerations of these two objects have the same magnitude but opposite directions. (Assume no other forces act on either object.)
Hint: What if the two objects have different mass?
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False.
Newton's 3rd law can be summarized as follows: A physical interaction (e.g., gravity) operates between two interacting objects and generates a pair of opposite forces, one on each object. It offers you a way to test for real forces (i.e., those that belong on the force side of ∑F⃗ =ma⃗ )—there should be a 3rd law pair force operating on some other object for each real force that acts on the object whose acceleration is under consideration.
Since F=ma: If the forces are equal in magnitude, must the accelerations also be of equal magnitude?
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According to Newton's 3rd law, the force on the (smaller) moon due to the (larger) earth is ______ in magnitude to, and in the ____ direction from, the force on the earth due to the moon.
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EQUAL
OPPOSITE
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⊗NEXT---->
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Fun With Spring Gun
A spring-loaded toy gun is used to shoot a ball of mass m=1.50kg straight up in the air, as shown in the figure. (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.
Find vm the muzzle velocity of the ball (i.e., the velocity of the ball at the spring's equilibrium position y=0).
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Find the spring potential energy first. The spring is compressed 25.0cm which is 0.250m:
U = 1/2kx²
U = 1/2(667)(0.250)²
U = 20.84375J
Now find the ball's velocity when it's released - just solve for kinetic energy:
KE = 1/2mv2²
20.84375 = 1/2(1.50)v²
20.84375 = 0.75v²
27.79167 = v²
v = √27.8
v = 5.27m/s
Next, find the velocity at y = 0. Since the spring was compressed 0.250m, that means the initial velocity is at y = -0.250 and the ball will travel 0.250m upwards to reach y = 0. We can use the kinematic equation:
vf² = v0² + 2α(d)
α will be acceleration due to gravity, so:
vf²= 5.27² + 2(-9.8)(0.250)
vf² = 27.77 - 19.6(0.250)
vf² = 22.87
vf = 4.78m/s
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Find the maximum height hmax of the ball.
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We can solve this using the same kinematic equation:
vƒ² = v₀² + 2a(d).
****At the peak (maximum height), we know that the velocity will be zero.***
So: vƒ² = 0
v₀² -> the kinetic energy from above ==> 5.27
STEP 1
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vƒ² = v₀² + 2a(d)
0² = 5.27² - 19.6*d
0 = 27.77 - 19.6*d
19.6d = 27.77
d = 1.42m
STEP 2
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Subtract 0.250m since the ball started from y = -0.250:
1.42 - 0.250 = 1.17
= 1.17m
Note that the initial velocity above was from the moment the ball was released (y = -0.250). We could have also solved using the velocity at y = 0 and then wouldn't have had to subtract the 0.250m.
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Power and efficiency
Power is defined as the rate of doing work
Power = work done/time taken
SI unit is watt (W)
Efficiency is the ratio of useful output energy to the total input energy or the ratio of useful power to the total input power.
Efficiency = (useful output energy / input energy) x 100%
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What power is needed to lift a 344kg crate of bricks from the ground to the top of a 23.8m -high building in 1 minute?
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(Mass * Gravity * Height) ÷ Time (s)
= (344 * 9.8 * 23.8) ÷ 60
= 1337
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A rock is dropped from a 20m -high ledge.
What's its speed when it hits the ground?
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Height = (v²/2g)
20 = v² ÷ 2g
20*2*9.8 = v²
v = √392
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What's its height when its speed is half the value found in previous part?
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We now apply (1) again but this time let Vf = 1/2 Vi
so that h is the height at which V is one half
the inital value. Then we have
(4) (1/2Vi)² -Vi² = -2gh
or
(5) -3/4 Vi² = -2gh
or solving for h
(6) 3/8 Vi² ÷ 9.8 ≈ 15m
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Hooke :
One of the pioneers of modern science, Sir Robert Hooke (1635-1703), studied the elastic properties of springs and formulated the law that bears his name. Hooke found the relationship among the force a spring exerts, , the distance from equilibrium the end of the spring is displaced, , and a number called the spring constant (or, sometimes, the force constant of the spring). According to Hooke, the force of the spring is directly proportional to its displacement from equilibrium, or
F = -kx.
In its scalar form, this equation is simply
F = -kx.
The negative sign indicates that the force that the spring exerts and its displacement have opposite directions. The value of depends on the geometry and the material of the spring; it can be easily determined experimentally using this scalar equation.
Toy makers have always been interested in springs for the entertainment value of the motion they produce. One well-known application is a baby bouncer,which consists of a harness seat for a toddler, attached to a spring. The entire contraption hooks onto the top of a doorway. The idea is for the baby to hang in the seat with his or her feet just touching the ground so that a good push up will get the baby bouncing, providing potentially hours of entertainment.
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....
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What is the spring constant of the spring being tested for the baby bouncer?
Express your answer as a fraction in unsimplified form.
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Remember that F = -kx. All you have to do is rearrange the equation:
F = -kx
-k = F/x
Drop the minus sign since "k" is always positive:
k = F/x
So you can take any pair of values from the table and find k, but since mastering physics wants a simplified answer in fractional form, use:
k = 2.5/0.005
k = 2.5/0.005
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One of the greatest difficulties with setting up the baby bouncer is determining the right height above the floor so that the child can push off and bounce. Knowledge of physics can be really helpful here.
If the spring constant k = 5.0 * 10² N, the baby has a mass m = 11 kg, and the baby's legs reach a distance d = 0.15 m from the bouncer, what should be the height of the "empty" bouncer above the floor?
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For this problem, just find the force of gravity on the baby and set it equal to the force on the spring. Start with gravity:
Fg = mg
Fg = 11 * 9.8
Fg = 107.8 N
Now plug this value into the formula for the force of the spring.
F = -kx
107.8 = -(500) * x
x = -0.2156
Subtract the length of the baby's legs:
x = -0.2156 - 0.15
x = -0.3656
This is a negative value because the spring will be stretching *down* - so we know that the spring has to be 0.3656 m above the ground so that the baby's feet will just touch the floor.
= 0.37 m
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A particular spring has a force constant of 1.5×103 N/m.
How much work is done in stretching the relaxed spring by 5.0 cm?
Express your answer using two significant figures.
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Spring potential energy is given by U = 1/2kx2. When compressing or stretching an ideal spring, all of the work done to compress/stretch it is stored as potential energy. So:
U = 1/2kx2
U = 1/2(1500)(0.05)2
U = 1.9J
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How much more work is done in stretching the spring an additional 3.0 cm?
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Note that finding the potential energy for a stretch of 0.03m won't give us the right answer here - since x is squared, we need to take into consideration all of the stretch/compression that was already done. So we can solve for the potential energy after 0.08m and then subtract the potential energy from the first 0.05m:
U = 1/2kx2
U = 1/2(1500)(0.08)2
U = 4.8J
Now subtract the 1.9J from Part A:
4.8 - 1.9 = 2.9
So the answer is 2.9J
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Principle of Conservation of Energy
States that energy can neither be created not destroyed but can be transformed from one form into another with no change in its total amount.
Eg. A ball of mass 3kg is dropped from a height of 5m.
i. calculate the gravitational potential energy of the ball before it is dropped
ii. calculate the speed of the ball on hitting the ground
iii. if the ball bounces to a height of 3m, with what speed does it leave the ground?
iv. explain why the ball does not reach its original height when it bounces up again
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i. gravitational potential energy = mgh = 3 x 10 x 5 = 150J
ii. The kinetic energy of the ball on hitting the ground is equal to the ball's original gravitational potential energy so the kinetic energy of the ball on hitting the ground = 150J
If the ball hits the ground with speed v,
1/2 mv2 = 150
v2 = (150 - 2)/3 = 100
v = 10ms-1
iii. The kinetic energy of the ball on leaving the ground is equal to its gravitational potential energy on rising to its maximum height, that is 3m.
The gravitational potential energy of the ball 3m above the ground = 3 x 10 x 3 = 90 J
The kinetic energy of the ball leaving the ground = 90 J
If the ball leaves the ground with speed v,
1/2 mv2 = 90
v2 = (90 x 2)/3
v = 7.746ms-1
iv. Because part of its kinetic energy is changed into other forms of energy like sound and heat when it hits the ground
Eg. A pendulum bob of mass 0.5kg is moved sideways until it has risen by a vertical height of 0.2m. Calculate the speed of the bob at its
i. highest point
ii. lowest point
i. at the highest point, the kinetic energy of the bob = 0
if the speed of the bob at its highest point is v,
1/2 mv2 = 0
1/2 x 0.5 x v = 0
v2 = 0
v = 0
ii. according to the principle of conservation of energy, the kinetic energy of at the lowest point is equal to the gravitational potential energy at the highest point.
If the speed of the bob at its lowest point is v,
1/2 mv2 = mgh
v2 = 2 x 10 x 0.2 = 4
v = 2 m/s
