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Welcome back to Physics 211PowerPoint Presentation

Welcome back to Physics 211

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### Welcome back to Physics 211

Today’s agenda:

Review Exam 2

Potential energy of a spring

Conservation of mechanical energy

Conservation of total energy

Reminder …

- Professor Britton Plourde
- Office = 223 Physics Building
- Office hours = Tuesday, 2-4PM, or by appointment
- Next Mastering Physics homework set to be assigned Friday

1. [25 pts total] A block is at rest on a rough inclined plane as shown in the figure.

a. [5 pts] In the space below draw a free body diagram showing all forces on the block.

b. [5 pts] Find the components of all forces perpendicular to the plane. Hence write down an equation which ensures that the block does not accelerate in this direction.

c. [5 pts] Find the components of all forces along the plane. Hence write down an equation which ensures that the block does not accelerate down the plane.

1.d. [3 pts] If =300 and the mass of the block is 1 kg what is the magnitude of the friction force experienced by the block ? (take g=10 m/s2)

e. [7 pts] In the case described in part d what is the magnitude of the normal force experienced by the block. Use this value and the magnitude of the friction force computed in part d to calculate the minimum value for the coefficient of static friction necessary for the block to be at rest.

B

A

2. [25 pts total] A force P is applied by a hand to two blocks which are in contact on a frictionless, horizontal table as shown in the figure. The blocks accelerate to the right. Block A has a smaller mass than block B

a. [5 pts] Draw a free body diagram for block A. Label your forces with 2 subscripts indicating which body the force acts on and which body is supplying the force.

B

A

2.b. [5 pts] Draw a free body diagram for block B. Label your forces as in part a.

c. [4 pts] Rank the magnitudes of the normal forces on the blocks due the ground. What are their directions ? Explain your answers.

d. [4 pts] Which block experiences the larger net force ? What is the direction of this net force ? Explain your answer.

B

A

2.e. [2 pts] Identify which forces are Newton third law pairs

f. [5 pts] Suppose that initially the mass of block A were half that of block B. If in a subsequent experiment the mass of block A were doubled by what factor would the acceleration change assuming the pushing force remained constant ?

100g

3. [ 25 pts total] The figure shows a cart on a horizontal, frictionless table. The cart is connected to a massless inextensible string which passes over a frictionless pulley to a block of mass 100 g. The system is released from rest. The block is observed to be falling with an acceleration of 0.5 m/s2. Assume g=10 m/s2

a. [6 pts] Draw a free body diagram for the weight. Use it to compute the tension force on the weight.

100g

3.b. [6 pts] Draw a free body diagram for the cart. What is the acceleration of the cart ? Compute the mass of the cart.

c. [5 pts] What is the speed of the falling weight after 2 seconds ? How far has it fallen at that point ?

d. [4 pts] How much work has been done on the cart by the tension force after 2 seconds. What is its sign ?

e. [4 pts]Compute the kinetic energy of the cart after 2 seconds. Are the results of parts d. and e, compatible with the work-kinetic energy theorem ?

crate

top view

4.[25 pts total] A crate rests on the floor of a merry-go-round that rotates counterclockwise at a constant rate. The crate does not slip on the floor of the merry-go-round. Assume g=10 m/s2.

a. [4 pts] In the top view diagram at right draw arrows representing the velocity and acceleration vector for the crate.

crate

top view

4.b. [4 pts] Draw a free body diagram for the crate pictured from a side view. Make sure you label the forces so as to show the object on which the force acts and the object exerting the force.

c. [5 pts] The mass of the crate is 4 kg and its speed relative to the Earth is 3.0 m/s. The radius of the circle through which the crate moves is 2.0 m and the coefficient of static friction between the box and merry-go-round is =0.8. Find the magnitude of the net force on the crate. Show your work.

crate

top view

4.d. [5 pts] The crate is now placed closer to the center of the merry-go-round, such that the radius of the circle through which it moves is 1.0 m. The merry-go-round turns at the same rate as before. What is the new net force on the crate ?

e. [7 pts] Now the crate is placed so that the radius of its path is 4.0 m. Will it slip ? Show your working

Total mechanical energy for object moving under gravity

- W-KE theorem now reads:
- D (K+Ug)=0 E=Ug+K=constant
- E is called the (mechanical) energy
- It is conserved
½ mv2+mgh = constant

Demo

Piledriver

- Ug of metal block converted to kinetic energy
- Kinetic energy converted to work to drive nail into wood
- Assume work required to drive nail is proportional to depth

Another example: springs …

Force F=-kx (Hooke’s law)

(x extension)

W=-1/2kx2

Therefore, can define elastic (spring) potential energy

U= 1/2kx2

Motion ?

- Imagine release from rest at x=x0 – what happens ?
1/2kx02=1/2kx2+1/2mv2 or

1/2mv2= 1/2kx02-1/2kx2

- what range does x lie in ?
- when is v greatest ?
- describe motion …

- A 0.5 kg mass is attached to a spring on a horizontal frictionless table. The mass is pulled to stretch the spring 5.0 cm and is released from rest. When the mass crosses the point at which the spring is not stretched, x=0, its speed is 20 cm/s. If the experiment is repeated with a 10.0 cm initial stretch, what speed will the mass have when it crosses x=0 ?
- 40 cm/s
- 0 cm/s
- 20 cm/s
- 10 cm/s

Many forces frictionless table. The mass is pulled to stretch the spring 5.0 cm and is released from rest. When the mass crosses the point at which the spring is not stretched, x=0, its speed is 20 cm/s. If the experiment is repeated with a 10.0 cm initial stretch, what speed will the mass have when it crosses x=0 ?

- For a particle which is subject to several (conservative) forces F1, F2 …
E=1/2mv2+U1+U2+ … is constant

- Principle called
Conservation of total mechanical energy

- A compressed spring fires a ping pong ball vertically upward. If the spring is compressed by 1 cm initially the ball reaches a height of 2 m above the spring. What height would the ball reach if the spring were compressed by just 0.5 cm ?
- (neglect air resistance)
- 2 m
- 1 m
- 0.5 m
- we do not have sufficient information to calculate the new height

Mass hanging on spring upward. If the spring is compressed by 1 cm initially the ball reaches a height of 2 m above the spring. What height would the ball reach if the spring were compressed by just 0.5 cm ?

xeq = mg/k

z = x + xeq

(1/2)mv2 + mgx + (1/2)kx2 =

(1/2)mv2 + (1/2)kz2 - (1/2)(mg)2/k

Mass hanging on spring upward. If the spring is compressed by 1 cm initially the ball reaches a height of 2 m above the spring. What height would the ball reach if the spring were compressed by just 0.5 cm ?

oscillations between z = -x0 and z = x0

Demo

Oscillations of vertical spring and mass

Many particles upward. If the spring is compressed by 1 cm initially the ball reaches a height of 2 m above the spring. What height would the ball reach if the spring were compressed by just 0.5 cm ?

- When system consists of many particles it is only the sum of all the particles energies which remains constant

Summary upward. If the spring is compressed by 1 cm initially the ball reaches a height of 2 m above the spring. What height would the ball reach if the spring were compressed by just 0.5 cm ?

- Total (mechanical) energy of an isolated system is constant in time.
- Must be no non-conservative forces
- Must sum over all conservative forces
- Must sum over all particles making up system

Nonconservative forces upward. If the spring is compressed by 1 cm initially the ball reaches a height of 2 m above the spring. What height would the ball reach if the spring were compressed by just 0.5 cm ?

- Can do work, but cannot be represented by a potential energy function
- Total mechanical energy can now change due to work done by nonconservative force
- For example, frictional force leads to decrease of total mechanical energy -- energy converted to heat, or internal energy
- Total energy = total mechanical energy + internal energy is conserved

Demo upward. If the spring is compressed by 1 cm initially the ball reaches a height of 2 m above the spring. What height would the ball reach if the spring were compressed by just 0.5 cm ?

Vertical spring and mass with damping

- Damping from air resistance
- Amplitude of oscillations decays
- Oscillations still about x = -xeq

Spring again upward. If the spring is compressed by 1 cm initially the ball reaches a height of 2 m above the spring. What height would the ball reach if the spring were compressed by just 0.5 cm ?

oscillates!

motion confined to

region below dotted

line

U(x)=1/2kx2

-a

a

x

Force as slope of potential energy graph U vs x upward. If the spring is compressed by 1 cm initially the ball reaches a height of 2 m above the spring. What height would the ball reach if the spring were compressed by just 0.5 cm ?

Definition : DU=-FDx

or F= - DU/Dx

Force proportional to slope of U(x)

curve

Equilibrium corresponds to F=0

ie zero slope

Potential Energy graphs upward. If the spring is compressed by 1 cm initially the ball reaches a height of 2 m above the spring. What height would the ball reach if the spring were compressed by just 0.5 cm ?

unstable equilibrium

U(x)

E

stable equilibrium

true equilibrium

x

Exam 2 upward. If the spring is compressed by 1 cm initially the ball reaches a height of 2 m above the spring. What height would the ball reach if the spring were compressed by just 0.5 cm ?

- Exams graded !
- Mean score 68.0/100.0 B-
- Score 51-67 (C- C+)
- Score 68-84 (B- B+)
- Score 85-92 A-, 93+ A

- Exams solutions will be posted online

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