11) boxes a and b are stacked on top of a frictionless floor as shown in the diagram. the coefficient of static friction between the boxes is 0.7. box a has mass 10 kg, and box b has mass 8 kg. a string is attached to box a, and a horizontal force f is applied to the string. a. what is the maximum value of f such that box b wouldn’t slip off of box a? the scenario is then repeated, but with a new box a that has a greater mass than the first one. b. how does the maximum value of f for this scenario compare to your answer to part a? justify your answer. c. how does the maximum acceleration for this scenario compare to your answer to part b? justify your answer. 12) two blocks made of the same material but with different masses m and 2m are placed on identical ramps as shown below. both blocks are released from rest from the same height up the ramp at the same time and accelerate down the ramps. friction is not negligible. two physics students are making predictions about the outcome of the scenario. student 1 says: “the blocks will reach the bottom of the ramp at the same time. the acceleration down the incline due to the parallel component of the force of gravity isn’t dependent on mass. since the blocks are made of the same material, they have the same force of friction acting on them.” student 2 says: “the block with mass 2m will reach the bottom first. the block with mass 2m has more mass, so the force of gravity acting on it will be greater. if it has a greater force of gravity acting on it, its acceleration will be greater than the other block.” a.) which aspects of student 1’s reasoning, if any are correct? explain. b.) which aspects of student 1’s reasoning, if any are incorrect? explain. c.) which aspects of student 2’s reasoning, if any are correct? explain. d.) which aspects of student 2’s reasoning, if any are incorrect? explain. e.) conclude which block will reach the end of the ramp 1st. justify your answer. include any relevant equations in your justification. 13) the graph shows the frictional force vs. the applied force on an object over time. the object is on a flat, horizontal surface. these are the only two horizontal forces on the object. the mass of the object is 15kg. a) draw two free body diagrams for the object. one for any time between a and b and one for c. b) describe the motion of the object from a to b and from b to c. cer c) sketch a graph of the velocity of the object vs. time for a to c. d) what must be the slope of the line from a to b (on the $f_f$ vs $f_a$ graph)? cer e) calculate the coefficient of static friction between the object and the surface. show all work. f) once the object is moving, how much applied force is needed to keep the object at constant velocity? justify your answer with cer. g) calculate the coefficient of kinetic friction between the object and the surface. show all work. h) the mass of the object is doubled. sketch a graph with the same axes and describe how the graph has changed or not changed. justify your answer with cer. i) the coefficient of static and kinetic friction are both doubled (back to original mass of 15kg). sketch a graph with the same axes and describe how the graph has changed or not changed. justify your answer with cer.

Question

Accepted Answer

### Problem 11a
# Explanation:
## Step 1: Analyze forces on Box B
For Box B not to slip, the static friction force provides the acceleration. The maximum static friction is $ f_s = \mu_s N $, where $ N = m_B g $ (normal force on B). So $ f_s = \mu_s m_B g $, and this equals $ m_B a $ (Newton's second law for B), so $ a = \mu_s g $.

## Step 2: Analyze forces on the system (A + B)
The total mass is $ M = m_A + m_B $. The applied force $ F $ causes acceleration $ a $, so $ F = M a = (m_A + m_B)\mu_s g $.

## Step 3: Substitute values
$ \mu_s = 0.7 $, $ m_A = 10 \, \text{kg} $, $ m_B = 8 \, \text{kg} $, $ g = 9.8 \, \text{m/s}^2 $.

$ F = (10 + 8) \times 0.7 \times 9.8 $

$ F = 18 \times 0.7 \times 9.8 $

$ F = 18 \times 6.86 $

$ F = 123.48 \, \text{N} $

# Answer:
The maximum value of $ F $ is $ \boldsymbol{123.48 \, \text{N}} $ (or approximately $ 123 \, \text{N} $ or $ 126 \, \text{N} $ if $ g = 10 \, \text{m/s}^2 $ is used).

### Problem 11b
# Brief Explanations:
When $ m_A $ increases (new $ m_A' > m_A $), the total mass $ M' = m_A' + m_B $ increases. From $ F = (m_A + m_B)\mu_s g $, since $ \mu_s $, $ g $, and $ m_B $ are constant, and $ m_A' > m_A $, $ M' > M $, so the new maximum $ F' $ is greater than the original $ F $.

# Answer:
The maximum $ F $ in the new scenario is greater than the answer in part a. This is because $ F = (m_A + m_B)\mu_s g $, and increasing $ m_A $ increases the total mass, leading to a larger $ F $ for the same acceleration (determined by Box B's friction).

### Problem 11c
# Brief Explanations:
The acceleration of the system is $ a = \mu_s g $ (from Box B's analysis: $ a = \frac{f_s}{m_B} = \mu_s g $). This acceleration depends only on $ \mu_s $ and $ g $, which are constant. So the maximum acceleration remains the same as in part a (and thus related to part b's $ F $ change, but acceleration itself is unchanged).

# Answer:
The maximum acceleration remains the same as in part a (and thus compared to part b's $ F $ increase, acceleration is constant). This is because $ a = \mu_s g $, independent of the mass of A (as long as B doesn't slip, acceleration is determined by static friction on B, which depends only on $ \mu_s $ and $ g $).

### Problem 12a
# Brief Explanations:
Student 1 is correct that the parallel component of gravity ( $ mg \sin\theta $) is independent of mass (since $ a_{\parallel} $ from gravity alone would be $ g \sin\theta $, mass cancels). Also, since the blocks are same material, $ \mu $ is same, but friction force is $ f = \mu N = \mu mg \cos\theta $, which depends on mass. Wait, no—Student 1 says "same force of friction", which is incorrect, but the first part about acceleration from parallel gravity being mass - independent is correct. So the aspect about the parallel component of gravity not depending on mass is correct.

# Answer:
The aspect that the acceleration down the incline due to the parallel component of gravity isn't dependent on mass is correct. This is because the parallel component of gravity is $ F_{\parallel} = mg \sin\theta $, and by Newton's second law $ F_{\parallel}-f = ma $, but the "acceleration due to the parallel component alone" (ignoring friction for a moment) would be $ a = g \sin\theta $, which is mass - independent.

### Problem 12b
# Brief Explanations:
Student 1 is incorrect in saying the blocks have the same force of friction. Friction force is $ f=\mu N=\mu mg \cos\theta $, so the block with mass $ 2m $ has twice the friction force (since $ f \propto m $) as the block with mass $ m $.

# Answer:
The aspect that the blocks have the same force of friction is incorrect. Friction force is $ f = \mu N $, and $ N = mg \cos\theta $, so $ f=\mu mg \cos\theta $. Thus, the block with mass $ 2m $ has a friction force of $ \mu(2m)g \cos\theta $, while the block with mass $ m $ has $ \mu mg \cos\theta $, so they are not the same.

### Problem 12c
# Brief Explanations:
Student 2 is correct that the block with mass $ 2m $ has a greater force of gravity ( $ F_g = mg $, so $ F_{g,2m}=2mg $ is greater than $ F_{g,m}=mg $).

# Answer:
The aspect that the block with mass $ 2m $ has a greater force of gravity acting on it is correct. The force of gravity (weight) is $ F_g = mg $, so for mass $ 2m $, $ F_g = 2mg $, which is greater than $ F_g = mg $ for mass $ m $.

### Problem 12d
# Brief Explanations:
Student 2 is incorrect in saying that greater gravity means greater acceleration. Using Newton's second law for a block on the ramp: $ F_{\parallel}-f = ma $, $ mg \sin\theta-\mu mg \cos\theta = ma $, dividing both sides by $ m $ gives $ a = g \sin\theta-\mu g \cos\theta $, which is independent of mass. So acceleration is the same for both blocks, so the claim that greater gravity (mass) leads to greater acceleration is wrong.

# Answer:
The aspect that a greater force of gravity leads to greater acceleration is incorrect. From $ mg \sin\theta-\mu mg \cos\theta = ma $, mass $ m $ cancels, giving $ a = g(\sin\theta - \mu \cos\theta) $, which is independent of mass. Thus, both blocks have the same acceleration, regardless of their mass.

### Problem 12e
# Explanation:
## Step 1: Recall the acceleration formula
For a block on an inclined plane with friction, the acceleration is $ a = g \sin\theta-\mu g \cos\theta $. As shown in part d, this acceleration is independent of mass.

## Step 2: Analyze motion time
The distance down the ramp $ d $ is related to initial velocity $ v_0 = 0 $, acceleration $ a $, and time $ t $ by $ d = v_0 t+\frac{1}{2}at^2 $, so $ t=\sqrt{\frac{2d}{a}} $.

Since both blocks have the same acceleration $ a $ (from part d) and start from the same height (so same $ d $ along the ramp, as $ d=\frac{h}{\sin\theta} $ where $ h $ is height, same for both), they will have the same time $ t $ to reach the bottom.

# Answer:
Both blocks will reach the bottom at the same time. The acceleration of each block is $ a = g(\sin\theta - \mu \cos\theta) $, which is independent of mass. Using the kinematic equation $ d=\frac{1}{2}at^2 $ (with $ v_0 = 0 $), and since both blocks travel the same distance $ d $ (same height, same ramp angle) with the same acceleration $ a $, they will take the same time $ t=\sqrt{\frac{2d}{a}} $ to reach the bottom.

### Problem 13e
# Explanation:
## Step 1: Identify maximum static friction
From the graph, the maximum static friction $ f_{s,\text{max}} $ is the value at point B, which is $ 10 \, \text{N} $ (assuming the graph shows $ f_s $ vs $ F_A $, and at B, $ f_s $ reaches maximum).

## Step 2: Normal force on the object
The normal force $ N = mg $, where $ m = 15 \, \text{kg} $, $ g = 9.8 \, \text{m/s}^2 $. So $ N = 15 \times 9.8 = 147 \, \text{N} $.

## Step 3: Coefficient of static friction formula
$ \mu_s=\frac{f_{s,\text{max}}}{N} $

## Step 4: Substitute values
$ \mu_s=\frac{10}{147}\approx0.068 $ (if $ f_{s,\text{max}} = 10 \, \text{N} $) or if the graph's maximum static friction is, say, $ 12 \, \text{N} $ (depending on the graph's scale—assuming the grid is 2 N per line, so 10 N? Wait, the graph has a mark at 10 and 8? Maybe the maximum static friction is $ 12 \, \text{N} $? Wait, the user's graph: "friction, N" on y - axis, with 10 and 8 marked. Let's assume $ f_{s,\text{max}} = 12 \, \text{N} $ (maybe the peak is at 12). Wait, no, the original problem: "the graph shows the frictional force vs. the applied force... mass of the object is 15kg". Let's use $ f_{s,\text{max}} = 10 \, \text{N} $ (as per the graph's y - axis mark).

Wait, $ N = mg = 15\times9.8 = 147 \, \text{N} $

$ \mu_s=\frac{f_{s,\text{max}}}{N}=\frac{10}{147}\approx0.068 $ (or if $ f_{s,\text{max}} = 12 $, $ \mu_s=\frac{12}{147}\approx0.0816 $)

# Answer:
If $ f_{s,\text{max}} = 10 \, \text{N} $, $ \mu_s=\boldsymbol{\approx0.068} $; if $ f_{s,\text{max}} = 12 \, \text{N} $, $ \mu_s=\boldsymbol{\approx0.082} $ (depends on graph's peak value).

### Problem 13g
# Explanation:
## Step 1: Identify kinetic friction force
After point B, the frictional force is kinetic friction $ f_k $, which is constant (from B to C). From the graph, let's say $ f_k = 8 \, \text{N} $ (as per the y - axis mark).

## Step 2: Normal force is still $ N = mg = 15\times9.8 = 147 \, \text{N} $

## Step 3: Coefficient of kinetic friction formula
$ \mu_k=\frac{f_k}{N} $

## Step 4: Substitute values
$ \mu_k=\frac{8}{147}\approx0.054 $ (if $ f_k = 8 \, \text{N} $)

# Answer:
If $ f_k = 8 \, \text{N} $, the coefficient of kinetic friction is $ \boldsymbol{\approx0.054} $.

(Note: For other sub - questions, like drawing free - body diagrams, describing motion, sketching graphs, etc., the answers would involve:

- Free - body diagrams: For A - B (static friction), forces on object are gravity (down), normal (up), applied force (right), static friction (left, equal to applied force). For C (kinetic friction), forces are gravity (down), normal (up), applied force (right), kinetic friction (left, constant, less than maximum static friction).

- Motion from A - B: Object is at rest or moving with constant velocity? No, from A to B, applied force increases, static friction increases to match it, so acceleration is zero (net force zero), so object is stationary (or moving at constant velocity if initially moving? Wait, no—static friction: if applied force is less than maximum static friction, and object is at rest, it remains at rest. So from A to B, net force is zero, so object is stationary (acceleration zero). From B to C, applied force is greater than maximum static friction, so kinetic friction acts, net force is $ F_A - f_k $, so object accelerates (since $ F_A $ may be constant or increasing? Wait, the graph is $ f_F $ vs $ F_A $. From A to B: $ f_F = F_A $ (static friction), so net force $ F_{net}=F_A - f_F = 0 $, so object is at rest (or moving with constant velocity if it was already moving, but initially at rest). At B, $ f_F $ reaches maximum static friction. After B, $ f_F = f_k $ (constant), so if $ F_A $ continues to increase, net force $ F_A - f_k $ is positive, so object accelerates.

- Velocity vs time graph: From A to B, velocity is zero (object at rest). At B, when $ F_A $ exceeds $ f_{s,\text{max}} $, velocity starts to increase, and since $ F_A - f_k $ may be constant (if $ F_A $ is constant after B), velocity increases linearly (constant acceleration) from B to C.

- Slope from A to B: The line from A to B has slope 1, because $ f_F = F_A $ (static friction equals applied force), so $ \frac{\Delta f_F}{\Delta F_A}=1 $.

- Applied force for constant velocity: Once moving, kinetic friction acts, so to have constant velocity, net force is zero, so $ F_A = f_k $. From the graph, $ f_k $ is the constant frictional force after B, so $ F_A = f_k $ (e.g., 8 N if $ f_k = 8 \, \text{N} $).

- Mass doubled: The normal force doubles ($ N' = 2mg $), so maximum static friction $ f_{s,\text{max}}'=\mu_s N' = 2\mu_s N = 2f_{s,\text{max}} $, and kinetic friction $ f_k'=\mu_k N' = 2f_k $. So the graph's A - B line will have the same slope (1), but the maximum static friction and kinetic friction values will double (the y - axis values for B and C will be twice as much).

- Coefficients doubled: Maximum static friction $ f_{s,\text{max}}'=\mu_s' N = 2\mu_s N = 2f_{s,\text{max}} $, kinetic friction $ f_k'=\mu_k' N = 2\mu_k N = 2f_k $. The slope from A to B remains 1, but the peak (B) and the kinetic friction (C) values double. The applied force at which slipping occurs (B's x - coordinate) will increase? No, $ f_{s,\text{max}}=\mu_s N $, so if $ \mu_s $ doubles, $ f_{s,\text{max}} $ doubles, so the applied force at B (where $ F_A = f_{s,\text{max}} $) will double. So the graph's A - B line goes to a higher y - value (double $ f_{s,\text{max}} $) at a higher x - value (double $ F_A $ at B), and the C line is at double $ f_k $.)

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QUESTION IMAGE

Question

Explanation:

Problem 11a

Step 1: Analyze forces on Box B

Step 2: Analyze forces on the system (A + B)

Step 3: Substitute values

Answer:

Problem 11b