AP Physics C - Mechanics Energy Problem Solving Techniques - - PDF document

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AP Physics C - Mechanics

Energy Problem Solving Techniques

2015-12-03 www.njctl.org

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· Gravitational Potential Energy Problem Solving · GPE, KE and EPE Problem Solving · Conservation of Energy Problem Solving · Nonlinear Spring · The Spring and the Roller Coaster · Potential Energy Graph Interpretation · Introduction

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Introduction

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Slide 5 / 84 Introduction

This is not a typical chapter presentation. It is a mix of step by step energy problem solutions, mixed in with free response and multiple choice formative assessment questions. The first four problems are non-calculus based. The remaining problems require calculus. These can be done in class, led by the teacher, or they can be done by the students outside of class.

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Gravitational Potential Energy Problem Solving

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Slide 7 / 84 GPE on an Incline

h d # Let's put together the concepts of two dimensional motion and forces with GPE. We'll use a box being pushed up an incline. It all depends on what we can

• measure. Assume it's easier to

measure the displacement (d) the box travels. How do we find its change in GPE?

Slide 8 / 84 GPE on an Incline

h d # The box starts with no velocity, and after it is pushed up a displacement d, the block slides up, and it momentarily stops before sliding back down. Does ΔGPE = mgd?

Slide 9 / 84 GPE on an Incline

h d # No! The formula for ΔGPE was calculated from the work formula, and it assumed the gravitational force (or the force that opposed it to lift the object) was in the same direction of the object's motion. The gravitational force points

• down. Since work only includes the

distance and force components that are in parallel, ΔGPE involves h and not d in the picture. How is h calculated from d?

Slide 10 / 84 GPE on an Incline

h d # ΔGPE = mgh When motion is along an incline, the change in height can be related to the distance traveled using trigonometry. sinθ = h/d h = dsin#

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1 A 5.0 kg block is at the top of a 6.0 m long frictionless ramp, which is at an angle of 370. What is the height of the ramp? 37o

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1 A 5.0 kg block is at the top of a 6.0 m long frictionless ramp, which is at an angle of 370. What is the height of the ramp? 37o

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Answer

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2 The 5.0 kg block slides to the bottom of the 6.0 m long frictionless ramp, which is at an angle of 37o. What is the change in its GPE? 37o

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2 The 5.0 kg block slides to the bottom of the 6.0 m long frictionless ramp, which is at an angle of 37o. What is the change in its GPE? 37o

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Answer

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GPE, KE and EPE Problem Solving

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Slide 14 / 84 Slide 15 / 84 GPE, KE and EPE

No, it does not! ΔGPE = mgh, where h is the vertical displacement (purely along the y axis) that the object has moved. The incline displacement is not important - only the vertical displacement. What about KE?

Slide 16 / 84 GPE, KE and EPE

As with all energy, KE is a scalar. However, it relates directly to the velocity, and velocity is a vector. When we perform calculations of velocity from KE and GPE, we need to be careful to relate the change in GPE

• nly to the change in KE in the y direction - thus it only

affects the velocity in the y direction. Now, what about EPE?

Slide 17 / 84 GPE, KE and EPE

In the case of EPE, the amount that the spring is compressed is the important variable - no trigonometry is required. Kinetic Energy will either use the vertical displacement an

• bject covers (for its relationship to GPE) or the actual

displacement of the object from the spring's force (for its relationship to EPE). Let's work a couple of problems by using the Conservation

• f TME to make this more clear.

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3 A projectile is fired at an angle of 450. Which factor is required to calculate the maximum height the projectile reaches by using the Conservation of Total Mechanical Energy? A The total initial velocity of the projectile. B The horizontal distance traveled by the projectile. C The total distance traveled by the projectile. D The x component of the velocity of the projectile. E The y component of the velocity of the projectile.

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3 A projectile is fired at an angle of 450. Which factor is required to calculate the maximum height the projectile reaches by using the Conservation of Total Mechanical Energy? A The total initial velocity of the projectile. B The horizontal distance traveled by the projectile. C The total distance traveled by the projectile. D The x component of the velocity of the projectile. E The y component of the velocity of the projectile.

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Answer E

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4 A spring launcher fires a marble at an angle of 52 to the

• horizontal. In calculating the energy available for

transformation into GPE and KE, what value of x is used in EPE = 1/2kx2 ?

Students type their answers here

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4 A spring launcher fires a marble at an angle of 52 to the

• horizontal. In calculating the energy available for

transformation into GPE and KE, what value of x is used in EPE = 1/2kx2 ?

Students type their answers here

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Answer The straight line displacement

• f the spring, independent of

the x and y axes.

Slide 20 / 84 Energy Problem Solving

What is the final velocity of a box of mass 5.0 kg that slides 6.0 m down a frictionless incline at an angle of 42 0 to the horizontal?

vo = 0 v = ?

The system will be the block. Since there is no friction, there are no external non conservative forces and we can use the Conservation of Total Mechanical Energy. What types

• f energy are involved here?

Slide 21 / 84 Energy Problem Solving

Only three types of energy have been discussed so far. And in this case, there is only GPE and KE:

vo = 0

d=6.0m θ=420 m = 5.0 kg

v = ?

(KE + EPE +GPE)0 = (KE + EPE + GPE) becomes: (KE + GPE)0 = (KE +GPE) to streamline the notation, we'll assume that no subscript implies a final quantity (KE = KEf)

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vo = 0

d=6.0m θ=420 m = 5.0 kg

v = ? h0=4.0m

The velocity at the bottom of the incline is 8.9 m/s.

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Consider the inclined plane problem that was just worked, but add a spring at the bottom

• f the incline.

The spring will be compressed a distance Δx and then released. Find the velocity of the box when it rises back to where it was first compressed - a height of Δh. What energies do we have to consider?

Slide 25 / 84 Energy Problem Solving

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Once compressed, the box has EPE, GPE and zero KE. When it loses touch with the spring at Δh above its fully compressed point, it will have KE, GPE and zero EPE. (KE + EPE + GPE) 0 = (KPE + EPE + GPE) becomes: (EPE + GPE)0 = (KE + GPE)

Slide 26 / 84 Energy Problem Solving

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Before we proceed further with the solution, think how hard this problem would be to solve without using Conservation of Energy. Once the object is released and the spring starts moving away from its compressed state, the force is no longer constant - it will require mathematical integration (calculus) to solve. Free body diagrams are not the best way to find the velocity of the object.

Slide 27 / 84 Energy Problem Solving

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Let's put in the equations and rearrange them to solve for v.

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We now have the equation for the velocity when the block rises a vertical displacement

• f Δh.

But if we're only given Δx, how do we find Δh? Use trigonometry and recognize that Δh = Δxsinθ.

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5 A box on an inclined plane is in contact with a spring. The box is released, compressing the spring. For every increment Δx, the box moves down the incline, how much does its height, Δh, change? A Δ x2 B Δ xcos θ C Δ xsin θ D √Δ x E Δ xtan θ

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5 A box on an inclined plane is in contact with a spring. The box is released, compressing the spring. For every increment Δx, the box moves down the incline, how much does its height, Δh, change? A Δ x2 B Δ xcos θ C Δ xsin θ D √Δ x E Δ xtan θ

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Answer C

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6 A box is held on top of a spring on an inclined plane of angle θ = 310. The box is released, compressing the

• spring. If the spring moves 7.0 m down the plane, how

much does its height, Δh, change? A 7.0 m B 6.5 m C 6.0 m D 3.6 m E 3.0 m

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6 A box is held on top of a spring on an inclined plane of angle θ = 310. The box is released, compressing the

• spring. If the spring moves 7.0 m down the plane, how

much does its height, Δh, change? A 7.0 m B 6.5 m C 6.0 m D 3.6 m E 3.0 m

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Answer D

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7 A box of mass m is on an inclined plane that makes an angle of θ with the horizontal and is in contact with a spring of spring constant k. The box is released, and it compresses the spring an amount Δx before rebounding. In terms of m, g, k and θ, what is the value of Δx?

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7 A box of mass m is on an inclined plane that makes an angle of θ with the horizontal and is in contact with a spring of spring constant k. The box is released, and it compresses the spring an amount Δx before rebounding. In terms of m, g, k and θ, what is the value of Δx?

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Answer

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8 A spring (k = 150 N/m) on an incline of θ = 540 is compressed a distance of Δx = .060 m along the incline by a mass of 0.042 kg and then released. What is its velocity when it passes the point where it was first compressed and loses touch with the spring?

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8 A spring (k = 150 N/m) on an incline of θ = 540 is compressed a distance of Δx = .060 m along the incline by a mass of 0.042 kg and then released. What is its velocity when it passes the point where it was first compressed and loses touch with the spring?

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Answer

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9 A marble launcher shoots a marble vertically and then shoots a marble in the horizontal direction. Describe why the exit velocity of the marble in the two cases is

• different. Which exit velocity is greater?

Students type their answers here

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9 A marble launcher shoots a marble vertically and then shoots a marble in the horizontal direction. Describe why the exit velocity of the marble in the two cases is

• different. Which exit velocity is greater?

Students type their answers here

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Answer In both cases, the EPE is the same. For the vertical case, part of the EPE is transformed into an increase in GPE as the exit point of the marble is higher than its compressed point. In the horizontal direction, there is no ΔGPE, hence the EPE is transformed totally into

• KE. The marble's velocity is greater

in the x direction.

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Conservation of Energy Problem Solving

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Slide 35 / 84 Falling Objects - the Energy way

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An object, at rest, falls from a height, h0, to the ground, and you want to find out what its velocity is right before it hits the ground (assume no air friction). Before you learned the Conservation of Energy, you would draw a free body diagram and then use a Kinematics equation to find the velocity. Review this with your group and then remove the screen below to check: mg

Slide 36 / 84 Falling Objects - the Energy way

Now, let's use the Conservation of Energy to solve this problem. Define the system as the object and GPE at the ground as zero. Since there is no friction, the net external non conservative work on the system will be zero (the force due to gravity is conservative and is taken into account by GPE). There is less algebra associated with the Conservation of TME approach then the dynamics and kinematics solution. The next chapter will present a problem where the Conservation

• f TME is much simpler than

using dynamics and kinematics.

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10 A ball of mass 0.45 kg falls from a building of height = 21 m. What is the ball's speed right before it hits the ground? A 14 m/s B 20 m/s C 210 m/s D 410 m/s

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10 A ball of mass 0.45 kg falls from a building of height = 21 m. What is the ball's speed right before it hits the ground? A 14 m/s B 20 m/s C 210 m/s D 410 m/s

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Answer B

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11 Two objects, one with a mass of 0.43 kg, and the other with a mass of 42.5 kg, fall from a height of 31.1 m. Which object has the greater velocity right before it hits the ground? (assume no friction) A Both have the same velocity. B The 0.43 kg object C The 42.5 kg object

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11 Two objects, one with a mass of 0.43 kg, and the other with a mass of 42.5 kg, fall from a height of 31.1 m. Which object has the greater velocity right before it hits the ground? (assume no friction) A Both have the same velocity. B The 0.43 kg object C The 42.5 kg object

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Answer A

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The Spring and the Roller Coaster

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Slide 40 / 84 The Spring and the Roller Coaster

Look at the below diagram. A block of mass m, is ejected by a compressed spring and spins around a couple of loops before exiting to the right. What is the velocity of the block just as it leaves the spring and at points A and B (assume no friction)? Could you use free body diagrams and Newton's Laws? A rA B rB

Slide 41 / 84 The Spring and the Roller Coaster

That would be very complicated! The net force on the roller coaster is constantly changing as it goes into, around and out of the loop due to the changing direction of the Normal force. The gravitational force is the same magnitude all around the loop and points down. But the Normal force is always pointing towards the center of the loop. The vector addition of both forces results in a constantly changing

• force. This would be very hard to work out.

But what about Conservation of TME? A rA B rB 1

Slide 42 / 84 The Spring and the Roller Coaster

How would Conservation of TME work? The system is the spring, the block and the loops. Assume no friction, hence there are no external non conservative forces on this system: E0 = E1 = EA = EB E0 is defined as the energy of the compressed spring right before it is released, E1 is the energy of the block right after it leaves the spring, and EA and EB are the energies at points A and B. A rA B rB 1

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A rA B rB 1 Intuitively, why is the velocity at point B less than point A? The block has a greater GPE at point B - so since TME is conserved, it has a smaller KE - hence, a smaller

• velocity. Note how we don't care how it got there - an

elliptical loop would give the same velocity.

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12 A spring gun, aimed in the horizontal direction with k = 250 N/m is compressed 0.05 m and released. How fast will a 0.025 kg dart go when it exits the gun?

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12 A spring gun, aimed in the horizontal direction with k = 250 N/m is compressed 0.05 m and released. How fast will a 0.025 kg dart go when it exits the gun?

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Answer

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13 A student uses a spring, with k = 180 N/m, to launch a marble vertically into the air. The mass of the marble is 0.0040 kg and the spring is compressed 0.030 m. How high will the marble go above its initially compressed position?

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13 A student uses a spring, with k = 180 N/m, to launch a marble vertically into the air. The mass of the marble is 0.0040 kg and the spring is compressed 0.030 m. How high will the marble go above its initially compressed position?

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Answer

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14 A student uses a spring gun (k = 120 N/m) to launch a marble at an angle of 520 to the horizontal (m = .0020 kg, Δx = 0.041 m). What is the maximum height that the marble will reach above its initially compressed position?

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14 A student uses a spring gun (k = 120 N/m) to launch a marble at an angle of 520 to the horizontal (m = .0020 kg, Δx = 0.041 m). What is the maximum height that the marble will reach above its initially compressed position?

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Answer

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15 A roller coaster car is pulled up to a height (A) of 50 m, where it then goes down the other side of the track. It traverses two other loops, one at a height of 40 m (B), and the second at a height of 30 m (C). Rank the velocities of the car at the three heights from greatest to least. A A > B > C B A > C > B C B > A > C D C > B > A E C > B > A F C > A > B

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15 A roller coaster car is pulled up to a height (A) of 50 m, where it then goes down the other side of the track. It traverses two other loops, one at a height of 40 m (B), and the second at a height of 30 m (C). Rank the velocities of the car at the three heights from greatest to least. A A > B > C B A > C > B C B > A > C D C > B > A E C > B > A F C > A > B

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Answer E

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16 Four objects are thrown with identical speeds in different directions from the top of a building. Which will be moving fastest when it strikes the ground? A B C D E All will have the same speed. h

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16 Four objects are thrown with identical speeds in different directions from the top of a building. Which will be moving fastest when it strikes the ground? A B C D E All will have the same speed. h

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Answer E

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17 Four objects are thrown with identical speeds in different directions from the top of a building. Which will hit the ground first? A B C D E All will hit at the same time. h

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17 Four objects are thrown with identical speeds in different directions from the top of a building. Which will hit the ground first? A B C D E All will hit at the same time. h

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Answer D

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18 Four objects are thrown with identical speeds in different directions from the top of a building. Which will go the highest? A B C D E All will reach the same height. h

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18 Four objects are thrown with identical speeds in different directions from the top of a building. Which will go the highest? A B C D E All will reach the same height. h

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Answer A

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19 Four objects are thrown with identical speeds in different directions from the top of a building. Which will land furthest from the base of the building? A B C D E All will land at the same place. h

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19 Four objects are thrown with identical speeds in different directions from the top of a building. Which will land furthest from the base of the building? A B C D E All will land at the same place. h

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Answer B

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20 Four objects are thrown with identical speeds in different directions from the top of a building. Which will have the greatest horizontal component of its velocity at its maximum height? A B C D E All will be the same. h

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20 Four objects are thrown with identical speeds in different directions from the top of a building. Which will have the greatest horizontal component of its velocity at its maximum height? A B C D E All will be the same. h

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Answer C

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21 Three objects are thrown with identical speeds in different directions from the top of a building. Which will have the greatest kinetic energy at its maximum height? A B C D All will have the same. h

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21 Three objects are thrown with identical speeds in different directions from the top of a building. Which will have the greatest kinetic energy at its maximum height? A B C D All will have the same. h

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Answer C

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Nonlinear Spring

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Slide 59 / 84 Nonlinear Spring

Hooke's Law shows the relationship between the force exerted by a spring on a mass to its displacement. The force for a perfect spring is linear (a function of the displacement x), and is opposite the displacement. Not all springs are linear. Some are functions of the displacement squared: For notation purposes, we will work with the magnitude of the force, and keep the sign indicating the force is opposite the displacement.

Slide 60 / 84 Nonlinear Spring

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This is the force done by the spring on the object stretching it. The force exerted by the object on the spring is: We will assume this is a conservative force. Why is it important to distinguish between the force exerted by the spring and the external force? What can be derived for a conservative force?

Slide 61 / 84 Nonlinear Spring

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If a system is associated with a conservative force, a potential energy can be calculated for it. When calculating the potential energy, you need to use the force from the field (gravitation) or the object (spring). Calculate the work done by the spring on an attached object when it is displaced from equilibrium x = 0 to xf = x. Calculate the potential energy stored in the system after it is displaced to point x.

Slide 62 / 84 Nonlinear Spring

Definition of Work for a non constant force The spring is stretched from x = 0 to x = x Substitution Integration Work done by the spring How is the potential energy calculated?

Slide 63 / 84 Nonlinear Spring

The potential energy is the negative of the work done by the field, or in this case, the spring. The potential energy is positive - which means that the work done by the external force increases the potential energy of the spring/mass system.

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22 What is the potential energy for a non linear spring where the force exerted by the spring on a mass is F(x) = -kx3? A kx4/4 B kx5/5 C -kx4/4 D -kx5/5 E kx2/2

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22 What is the potential energy for a non linear spring where the force exerted by the spring on a mass is F(x) = -kx3? A kx4/4 B kx5/5 C -kx4/4 D -kx5/5 E kx2/2

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Answer A

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23 What is the work done by a non linear spring on an attached mass where the force exerted by the spring is F(x) = -kx3? A kx4/4 B kx5/5 C -kx4/4 D -kx5/5 E kx2/2

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23 What is the work done by a non linear spring on an attached mass where the force exerted by the spring is F(x) = -kx3? A kx4/4 B kx5/5 C -kx4/4 D -kx5/5 E kx2/2

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Answer C

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24 What is the work done by a non linear spring on a mass where the force exerted by the spring on a mass is F(x) = -kx3? A kx4/4 B kx5/5 C -kx4/4 D -kx5/5 E kx2/2

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24 What is the work done by a non linear spring on a mass where the force exerted by the spring on a mass is F(x) = -kx3? A kx4/4 B kx5/5 C -kx4/4 D -kx5/5 E kx2/2

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Answer A

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Potential Energy Graph Interpretation

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Slide 68 / 84 Potential Energy Graph

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A system does not always have a simple potential energy, as multiple forces might be acting on it, with different magnitudes and directions, and range of effectiveness. Look up the Lennard-Jones potential in a textbook or on the web to see an example of a more complex potential energy model. Potential Energy vrs. Position graphs are used to show the dependence of the Potential Energy on position.

Slide 69 / 84 Potential Energy Graph

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Here is a sample potential energy - position graph. If you place a particle at point A, and it is at rest, what can you say about its total mechanical energy?

Slide 70 / 84 Potential Energy Graph

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Since TME = KE + U, the total mechanical energy of the system is equal to the potential energy at point A. Assume the particle is free to move under the influence of the potential energy. Will it move from point A, and if so, in which direction?

Slide 71 / 84 Potential Energy Graph

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In order to find its motion, the force due to the potential energy needs to be understood. We are just dealing with motion in the x direction: The derivative of U with respect to x is the slope of the line at any point. Since point A is the start of a straight line with negative slope, the force is positive - and the object will move to the right. If an object is released from rest at point A, how far in the x direction can it move?

Slide 72 / 84 Potential Energy Graph

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The TME of the system is 3 J. Due to the Conservation of Energy, the TME can never exceed 3 J. KE is always positive. At x = 7 m, the potential energy will be 3 J. This means KE = 0, and since TME cannot be greater than 3 J, the particle cannot move further than this point. Similarly, a particle released from rest at point B cannot go past point E.

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25 The potential energy as a function of position in an area

• f space is given by U(r) = 3r2 + 6r + 7. What is the force
• n a particle placed at r = 4.0 m?

A -30 N B -37 N C 0 N D 30 N E 37 N

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25 The potential energy as a function of position in an area

• f space is given by U(r) = 3r2 + 6r + 7. What is the force
• n a particle placed at r = 4.0 m?

A -30 N B -37 N C 0 N D 30 N E 37 N

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Answer A

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26 What is the force on an object placed at x = 0.5 m, and represented by the plot of potential energy versus position? A -6 N B -3 N C 0 N D 3 N E 6 N

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26 What is the force on an object placed at x = 0.5 m, and represented by the plot of potential energy versus position? A -6 N B -3 N C 0 N D 3 N E 6 N

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Answer D

Slide 75 / 84 Potential Energy Graph

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Release a particle of mass 1 x 10-2 kg from rest at point A. What will be its velocity at points B, C, D, E and F?

Slide 76 / 84 Potential Energy Graph

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UA = Ui + KEi where i = B, C, D, E, F.

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27 An object at rest is released from point A. Why does the

• bject not move past x = 6.9 m?

A The potential energy at that point is positive. B The potential energy is negative at that point. C The kinetic energy at x = 6.9 m is greater than the potential energy. D The kinetic energy at x = 6.9 m is positive. E The kinetic energy at x = 6.9 m is zero.

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27 An object at rest is released from point A. Why does the

• bject not move past x = 6.9 m?

A The potential energy at that point is positive. B The potential energy is negative at that point. C The kinetic energy at x = 6.9 m is greater than the potential energy. D The kinetic energy at x = 6.9 m is positive. E The kinetic energy at x = 6.9 m is zero.

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Answer E

Slide 78 / 84 Potential Energy Graph

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Where are the equilibrium points? State whether they are stable or unstable, and what would be the motion of a particle released at those points.

Slide 79 / 84 Potential Energy Graph

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Equilibrium points are where the slope of the potential energy - position curve is zero - which means the force

• n an object at that point is zero.

Points C, E and F are equilibrium points. What happens if an external force acts on objects at the equilibrium points?

Slide 80 / 84 Potential Energy Graph

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At points C and F, the object will move in the direction of the force. But in each case, a restoring force will act against the external force - this results in the object returning and passing its

• riginal position, where an opposite

restoring force will act to return it again. Will this continue forever?

Slide 81 / 84 Potential Energy Graph

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It depends. If an external force acts to the right at point F, and moves the object to x = 8 m, it will experience a restoring force to the

• left. It will have enough energy to

pass point F, and then continue past E, and rise to point A. In this example, we can't say what happens then, as the graph is cut

• ff.

If the force moves the object to x = 6.5 m, then it will oscillate about point F until friction stops it. When an object at an equilibrium point is moved and then returns to its

• riginal point, it is in

stable equilibrium.

Slide 82 / 84 Potential Energy Graph

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If an object at point E is disturbed slightly, in either direction, the force generated by the potential energy will support the displacement and move the object further from point E. When an object at an equilibrium point is moved and does not return to its

• riginal point, it is in

unstable equilibrium.

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28 A particle is released from rest at x = 1.0 m as presented by the below graph. Describe its subsequent motion, assuming no friction. A The particle will reach point E, and then return to point B. B The particle will come to rest at point C. C The particle will oscillate between x = 1 m and x = 3 m. D The particle will oscillate between x = 1.5 m and x = 2.5 m. E The particle will not pass point C.

Slide 83 (Answer) / 84

28 A particle is released from rest at x = 1.0 m as presented by the below graph. Describe its subsequent motion, assuming no friction. A The particle will reach point E, and then return to point B. B The particle will come to rest at point C. C The particle will oscillate between x = 1 m and x = 3 m. D The particle will oscillate between x = 1.5 m and x = 2.5 m. E The particle will not pass point C.

[This object is a pull tab]

Answer C

Slide 84 / 84

29 A particle is released from rest at point A as presented by the below graph. Describe its subsequent motion, assuming no friction. A The particle will reach point E, and then return to point A. B The particle will come to rest at point F. C The particle will oscillate between x = 1 m and x = 6.5 m. D The particle will oscillate between x = 0.5 m and x = 7.0 m. E The particle will come to rest at point C.

Slide 84 (Answer) / 84

29 A particle is released from rest at point A as presented by the below graph. Describe its subsequent motion, assuming no friction. A The particle will reach point E, and then return to point A. B The particle will come to rest at point F. C The particle will oscillate between x = 1 m and x = 6.5 m. D The particle will oscillate between x = 0.5 m and x = 7.0 m. E The particle will come to rest at point C.

[This object is a pull tab]

Answer D