Practice Questions: Dynamics

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SECTION I: MULTIPLE CHOICE

Directions

Select the best answer for each of the following questions. Each question has four answer choices labeled A, B, C, and D. Choose the one alternative that best completes the statement or answers the question.

Question 1

What is the magnitude of the friction force acting on the block?

1. 0 N
2. 15 N
3. 16 N
4. 20 N

Question 2

Which of the following expressions best represents the net horizontal force on the object?

1. \( F_{\text{net}} = 20 \cos(30°) \)
2. \( F_{\text{net}} = 20 \sin(30°) \)
3. \( F_{\text{net}} = (5.0)(2.0) \)
4. \( F_{\text{net}} = 20 - (5.0)(9.8) \)

Question 3

What is the velocity of the object at \( t = 6 \) s?

1. 8 m/s
2. 16 m/s
3. 24 m/s
4. 32 m/s

Question 4

A box is pushed across a rough horizontal floor at constant velocity. Which of the following statements is true about the forces acting on the box?

1. The applied force is greater than the friction force.
2. The applied force equals the friction force.
3. The net force on the box is in the direction of motion.
4. The friction force is zero because the box is moving.

Question 5

If the block slides down the incline at constant velocity, which of the following equations must be true?

1. \( mg \sin\theta = f \)
2. \( mg \cos\theta = f \)
3. \( N = mg \)
4. \( N = mg \sin\theta \)

Question 6

When the system is released from rest, what is the magnitude of the acceleration of the blocks?

1. 3.9 m/s²
2. 5.9 m/s²
3. 6.5 m/s²
4. 9.8 m/s²

Question 7

A student pushes a 1.5 kg book across a table with a force of 5.0 N. The book accelerates at 2.0 m/s². What is the magnitude of the friction force acting on the book?

1. 2.0 N
2. 3.0 N
3. 5.0 N
4. 7.0 N

Question 8

Which rope experiences the greater tension?

1. Rope A, because it makes a larger angle with the ceiling
2. Rope B, because it makes a smaller angle with the ceiling
3. Both ropes have equal tension because the object is in equilibrium
4. Cannot be determined without knowing the length of the ropes

Question 9

What is the coefficient of kinetic friction between the block and the carpet?

1. 0.41
2. 0.61
3. 0.82
4. 1.2

Question 10

A 50 kg person stands in an elevator that is accelerating upward at 2.0 m/s². What is the magnitude of the normal force exerted by the elevator floor on the person?

1. 390 N
2. 490 N
3. 590 N
4. 690 N

Question 11

What is the magnitude of the average friction force acting on the puck?

1. 0.26 N
2. 0.51 N
3. 1.3 N
4. 2.6 N

Question 12

If the only horizontal force acting on the cart is a constant applied force, and the cart's position is given by \( x = 1.5t^2 \) (where \( x \) is in meters and \( t \) is in seconds), what is the magnitude of the applied force?

1. 3.0 N
2. 6.0 N
3. 12 N
4. 24 N

Question 13

A 1200 kg car travels around a flat circular track of radius 50 m at a constant speed of 20 m/s. What is the magnitude of the net force acting on the car?

1. 480 N
2. 4800 N
3. 9600 N
4. 12000 N

Question 14

What is the magnitude of the object's acceleration?

1. 1.6 m/s²
2. 2.4 m/s²
3. 3.2 m/s²
4. 4.0 m/s²

Question 15

Based on the data, what is the magnitude of the friction force acting on the block?

1. 1.0 N
2. 2.0 N
3. 3.0 N
4. 6.0 N

Question 16

Two blocks, A and B, are placed in contact on a horizontal frictionless surface. Block A has mass 2.0 kg and block B has mass 3.0 kg. A horizontal force of 10 N is applied to block A, pushing both blocks to the right. What is the magnitude of the force that block A exerts on block B?

1. 2.0 N
2. 4.0 N
3. 6.0 N
4. 10 N

Question 17

If the sled moves up the hill at constant velocity, what is the approximate magnitude of the tension in the rope?

1. 21 N
2. 51 N
3. 71 N
4. 147 N

Question 18

What is the magnitude of the net force acting on the ball while it is in the air? (Ignore air resistance.)

1. 0.5 N
2. 4.9 N
3. 7.5 N
4. 15 N

Question 19

A 0.2 kg ball attached to a string swings in a horizontal circle of radius 0.8 m. The ball completes one revolution every 1.0 s. What is the magnitude of the tension in the string?

1. 1.6 N
2. 5.0 N
3. 6.3 N
4. 25 N

Question 20

When the skydiver is falling at terminal velocity, what is the magnitude of the air resistance force acting on the skydiver?

1. 0 N
2. 50 N
3. 300 N
4. 588 N

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SECTION II: FREE RESPONSE

Directions

Answer both of the following questions. Show all your work for each part of the question. The parts within a question may not have equal weight. You are encouraged to draw diagrams, graphs, or other visual representations to support your answers.

FRQ 1: Mathematical Routines (15 minutes)

1. Derive an expression for the magnitude of the acceleration of the block as it slides up the incline. Begin your derivation by writing a fundamental physics principle or equation. Express your answer in terms of \( g \), θ, and μ_k.
2. The block slides up the incline, momentarily comes to rest, and then slides back down. Derive an expression for the magnitude of the acceleration of the block as it slides down the incline. Express your answer in terms of \( g \), θ, and μ_k.
3. For the specific case where θ = 30° and μ_k = 0.20, calculate the ratio of the magnitude of the acceleration going up the incline to the magnitude of the acceleration going down the incline.
4. Explain qualitatively why the magnitude of the acceleration is different when the block moves up the incline compared to when it moves down the incline, even though the angle and coefficient of friction remain the same.

FRQ 2: Experimental Design and Analysis (20 minutes)

1. Describe an experimental procedure the student could use to determine μ_k. Your procedure should include:
   1. A labeled diagram showing the experimental setup
   2. The specific measurements to be taken
   3. How to use these measurements to calculate μ_k
2. The student decides to pull the block at constant velocity using the spring scale and record the scale reading (tension force) for different masses placed on top of the block. Describe how the student should graph the data to determine μ_k. Specifically:
   1. What quantity should be plotted on the vertical axis?
   2. What quantity should be plotted on the horizontal axis?
   3. What does the slope of the best-fit line represent?
3. The student collects the following data: On the axes below, plot the appropriate data points and draw a best-fit line.
   [Grid with vertical axis labeled "Tension (N)" from 0 to 5.0 and horizontal axis labeled "Normal Force (N)" from 0 to 16]
4. Using your graph from part (c), calculate the coefficient of kinetic friction μ_k between the block and the surface. Show your work clearly.
5. A student suggests that instead of pulling the block at constant velocity, the experiment could be done by placing the block on an inclined plane and adjusting the angle until the block slides down at constant velocity. Explain whether this alternative method would produce a valid measurement of μ_k, and justify your reasoning using physics principles.

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ANSWER KEY

Part A: Multiple Choice Answer Table

Part B: Free Response Detailed Answers

FRQ 1 - Answer Key

Part A: Derive acceleration going up the incline

Begin with Newton's second law:

\[ \sum F = ma \]

Draw a free-body diagram showing the block on the incline with:

• Weight \( mg \) acting downward
• Normal force \( N \) perpendicular to the incline
• Friction force \( f_k \) pointing down the incline (opposing motion up)

Choose a coordinate system with the x-axis parallel to the incline (positive up) and y-axis perpendicular to the incline.

Resolve weight into components:

• Component parallel to incline: \( mg\sin\theta \) (down the incline)
• Component perpendicular to incline: \( mg\cos\theta \) (into the incline)

Apply Newton's second law perpendicular to incline (y-direction):

\[ N - mg\cos\theta = 0 \] \[ N = mg\cos\theta \]

Kinetic friction force:

\[ f_k = \mu_k N = \mu_k mg\cos\theta \]

Apply Newton's second law parallel to incline (x-direction, taking down as negative):
When moving up, both gravity component and friction point down the incline:

\[ -mg\sin\theta - f_k = ma \] \[ -mg\sin\theta - \mu_k mg\cos\theta = ma \] \[ a = -g\sin\theta - \mu_k g\cos\theta \] \[ a = -g(\sin\theta + \mu_k\cos\theta) \]

Magnitude of acceleration going up:

\[ |a_{\text{up}}| = g(\sin\theta + \mu_k\cos\theta) \]

Part B: Derive acceleration going down the incline

Begin with Newton's second law:

\[ \sum F = ma \]

When the block slides down, the friction force points up the incline (opposing motion down).

Normal force remains the same:

\[ N = mg\cos\theta \] \[ f_k = \mu_k mg\cos\theta \]

Apply Newton's second law parallel to incline (taking down as positive):

\[ mg\sin\theta - f_k = ma \] \[ mg\sin\theta - \mu_k mg\cos\theta = ma \] \[ a = g\sin\theta - \mu_k g\cos\theta \] \[ a = g(\sin\theta - \mu_k\cos\theta) \]

Magnitude of acceleration going down:

\[ |a_{\text{down}}| = g(\sin\theta - \mu_k\cos\theta) \]

Part C: Calculate the ratio of accelerations

Given: θ = 30° and μ_k = 0.20

Calculate acceleration up:

\[ |a_{\text{up}}| = g(\sin 30° + 0.20\cos 30°) \] \[ |a_{\text{up}}| = g(0.50 + 0.20 \times 0.866) \] \[ |a_{\text{up}}| = g(0.50 + 0.173) \] \[ |a_{\text{up}}| = 0.673g \]

Calculate acceleration down:

\[ |a_{\text{down}}| = g(\sin 30° - 0.20\cos 30°) \] \[ |a_{\text{down}}| = g(0.50 - 0.173) \] \[ |a_{\text{down}}| = 0.327g \]

Calculate ratio:

\[ \frac{|a_{\text{up}}|}{|a_{\text{down}}|} = \frac{0.673g}{0.327g} = \frac{0.673}{0.327} = 2.06 \]

The ratio is approximately 2.1.

Part D: Qualitative explanation

The magnitude of the acceleration is different because friction always opposes the direction of motion. When the block moves up the incline, both the component of gravity parallel to the incline and the friction force point down the incline, so they add together to create a larger net force (and thus larger acceleration magnitude) opposing the motion. When the block moves down the incline, the friction force points up the incline while the gravity component points down, so friction partially cancels the gravitational force, resulting in a smaller net force and smaller acceleration magnitude. The friction force direction reverses with the direction of motion, causing the asymmetry in acceleration magnitudes.

FRQ 2 - Answer Key

Part A: Experimental procedure

Diagram: A labeled diagram should show:

• Wooden block on horizontal surface
• Spring scale attached horizontally to the block
• Arrow showing direction of pull (horizontal)
• Additional masses stacked on top of block (optional)

Procedure:

1. Measure and record the mass of the wooden block using a balance: \( m_{\text{block}} \)
2. Place the block on the horizontal surface
3. Attach the spring scale to the block horizontally
4. Pull the block at constant velocity across the surface while observing the spring scale
5. Record the tension force \( F_T \) shown on the spring scale when the block moves at constant velocity
6. Repeat steps 4-5 by adding known masses on top of the block and recording the total mass and corresponding tension for each trial

Calculation of μ_k:
When the block moves at constant velocity, the net force is zero, so:

\[ F_T = f_k \] \[ F_T = \mu_k N \] \[ F_T = \mu_k mg \] \[ \mu_k = \frac{F_T}{mg} \]

where \( m \) is the total mass (block + added masses) and \( g = 9.8 \, \text{m/s}^2 \).

Part B: Graphing procedure

i. Vertical axis: Tension force (N) or friction force (N)

ii. Horizontal axis: Normal force (N), which equals \( mg \) for the horizontal surface

iii. Slope representation: The slope of the best-fit line represents the coefficient of kinetic friction μ_k, because from \( f_k = \mu_k N \), plotting \( f_k \) versus \( N \) yields a linear relationship with slope μ_k.

Part C: Plot data and draw best-fit line

Students should plot the five data points:

• (4.9 N, 1.5 N)
• (7.4 N, 2.2 N)
• (9.8 N, 3.0 N)
• (12.3 N, 3.7 N)
• (14.7 N, 4.5 N)

The best-fit line should pass through or near the origin and have a positive slope, showing a linear relationship between normal force (horizontal) and tension/friction force (vertical).

Part D: Calculate coefficient of kinetic friction

Method: Calculate the slope of the best-fit line using two points on the line (preferably endpoints or points that clearly lie on the line).

Using the first and last data points:

\[ \text{slope} = \frac{\Delta F_T}{\Delta N} = \frac{4.5 \, \text{N} - 1.5 \, \text{N}}{14.7 \, \text{N} - 4.9 \, \text{N}} \] \[ \text{slope} = \frac{3.0 \, \text{N}}{9.8 \, \text{N}} = 0.306 \]

Alternatively, using a more precise calculation with intermediate points:

\[ \text{slope} = \frac{3.0 - 1.5}{9.8 - 4.9} = \frac{1.5}{4.9} = 0.306 \]

Or averaging several calculations:

From point 1 to point 3: \( \frac{3.0-1.5}{9.8-4.9} = \frac{1.5}{4.9} = 0.31 \)

From point 3 to point 5: \( \frac{4.5-3.0}{14.7-9.8} = \frac{1.5}{4.9} = 0.31 \)

Therefore, μ_k ≈ 0.31 (acceptable range: 0.29-0.32 depending on how students draw the best-fit line)

Part E: Evaluation of alternative method

Yes, this alternative method would produce a valid measurement of μ_k.

Justification: When the block slides down an inclined plane at constant velocity, the net force parallel to the incline is zero. This means the component of gravitational force down the incline equals the kinetic friction force up the incline:

\[ mg\sin\theta = f_k = \mu_k N = \mu_k mg\cos\theta \]

Dividing both sides by \( mg\cos\theta \):

\[ \tan\theta = \mu_k \]

Therefore, by measuring the angle θ at which the block slides at constant velocity, the student can calculate μ_k directly as \( \mu_k = \tan\theta \). This method is based on the same fundamental principle (equilibrium of forces when velocity is constant) and would yield an accurate value for the coefficient of kinetic friction.