AP Physics C Mechanics - NJCTLcontent.njctl.org/courses/science/ap-physics-c-mechanics/... · 2015. 12. 4. · AP Physics C Mechanics ... This is not a typical chapter presentation

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AP Physics C Mechanics Energy Problem Solving Techniques

20151203

www.njctl.org

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Table of ContentsClick on the topic to go to that section

• 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

Return to Tableof Contents

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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 noncalculus 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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GPE on an Incline

h d

θ

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?

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GPE on an Incline

h d

θ

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?

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GPE on an Incline

h d

θ

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?

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GPE on an Incline

h d

θ

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 = dsinθ

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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

Answ

er

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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

Answ

er

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GPE, KE and EPE Problem Solving


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GPE, KE and EPE

We can now solve problems with GPE, KE and EPE in two dimensions.

The key is understanding how these different forms of energy depend on their components along the x and y axis.

Let's look at GPE first. If an object is moving along an incline, does its potential energy depend on the displacement traveled along the incline?

Conservation of Total Mechanical Energy (TME)

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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?

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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 only to the change in KE in the y direction thus it only affects the velocity in the y direction.

Now, what about EPE?

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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 object 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 of 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.

Answ

er

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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 ?

Answ

er

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Energy Problem SolvingWhat 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 420 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 of energy are involved here?

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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

Let's put in the equations now:

Given (and using trig to find the value of h0 from d):

Find: v

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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 of 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?

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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)

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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.

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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 of 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 Δx2

B Δxcosθ

C Δxsinθ

D √Δx

E Δxtanθ

Answ

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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

Answ

er

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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?

Answ

er

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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?

Answ

er

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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?

Answ

er

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Conservation of Energy Problem Solving


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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

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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 of 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 Answ

er

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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 Answ

er

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The Spring and the Roller Coaster


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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

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The Spring and the Roller CoasterThat 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

rB10

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The Spring and the Roller CoasterHow 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

rB10

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E0 = E1 = Ea = Ef Assume the spring has a spring constant, k, was compressed Δx and released, and GPE = 0

along the base of the loops.

EPE = KE1 = (KE + GPE)A = (KE + GPE)B

Note how the height of points A and B is twice the radius of each loop.

A

rA

B

rB10

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What's nice about the Conservation of TME, and working out the algebra in advance, is we can solve for different quantities in the above equation depending on what we were given.

TME stays the same, no matter where the block is on its journey.

A

rA

B

rB10

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The Spring and the Roller CoasterFind the velocity at point A first:

Now we just need to substitute in the values. What else can we solve for by modifying this solution a little?

A

rA

B

rB10

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The Spring and the Roller CoasterThe velocity at point B is found the same way!

A

rA

B

rB10

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A

rA

B

rB10

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?

Answ

er

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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?

Answ

er

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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?

Answ

er

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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

Answ

er

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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

Answ

er

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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

Answ

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C

D

E All will hit at the same time.

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

Answ

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C

D

E All will reach the same height.

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

Answ

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C

D

E All will land at the same place.

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

Answ

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C

D

E All will be 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

Answ

er

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Nonlinear Spring


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Nonlinear SpringHooke'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.

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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?

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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.

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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?

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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

Answ

er

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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

Answ

er

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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

Answ

er

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Potential Energy Graph Interpretation


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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 LennardJones 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.

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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?

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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?

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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?

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

A 30 N

B 37 N

C 0 N

D 30 N

E 37 N

Answ

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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

Answ

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Release a particle of mass 1 x 102 kg from rest at point A. What will be its velocity at points B, C, D, E and F?

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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 object 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.

Answ

er

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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.

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

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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 original position, where an opposite restoring force will act to return it again.

Will this continue forever?

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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 off.

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 movedand then returns to its original point, it is in stable equilibrium.

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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 movedand does not return to its original 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.

Answ

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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.

Answ

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AP Physics C Mechanics - NJCTLcontent.njctl.org/courses/science/ap-physics-c-mechanics/... · 2015. 12. 4. · AP Physics C Mechanics ... This is not a typical chapter presentation