9) a 5 kg box is moving across on a horizontal conveyor belt in a factory. from the perspective of a stationary worker on the floor, the box is moving at a constant speed of 2 m/s to the right. the coefficient of kinetic friction between the box and surface is 0.3, and the coefficient of static friction is 0.5. for the following questions, include direction in your answers. a. what is the force of friction acting on the box during this time? justify your answer. b. the conveyor belt starts to speed up. what is the largest acceleration the belt can have without the box slipping? justify your answer. c. if the box reaches a speed of 6 m/s, and then the conveyor belt comes to an abrupt halt, how much time will it take the box to slide to a stop, assuming it doesnt hit anything else on the belt? how far does it slide? justify your answer. 10) 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 b, and a horizontal force f is applied to the string. a. what is the maximum value of f such that box b wouldnt slip off of box a? show your work. b. what is the maximum acceleration of the two - box system such that box b wouldnt slip off box a? the scenario is then repeated, but with a new box a that has a greater mass than the first one. c. how does the maximum value of f for this scenario compare to your answer to part a? justify your answer. d. how does the maximum acceleration for this scenario compare to your answer to part b? justify your answer.

Question

Accepted Answer

### Question 9a
# Explanation:
## Step1: Analyze motion state
The box moves at constant speed, so net force is zero.
## Step2: Identify friction type
Since it's moving, kinetic friction? Wait, no—constant speed means acceleration $a = 0$. Forces: gravity $mg$ down, normal $N = mg$ up, and friction. For horizontal motion, if speed is constant, the applied force (from conveyor) must balance friction. But wait, when moving at constant speed on a conveyor, is there a driving force? Wait, no—if the box is moving at same speed as conveyor, then relative to conveyor, it's stationary. Wait, maybe I messed up. Wait, the worker sees it moving at 2 m/s right, constant speed. So acceleration $a = 0$. So net force $F_{net}=ma = 0$. The forces on the box: weight $mg$ down, normal $N = mg$ up, and friction. But if it's moving at constant speed, the friction force must be equal and opposite to any applied force. But in this case, the conveyor belt is moving, so the box is moving with it. Wait, maybe the box is in equilibrium. So horizontal forces: if there's no acceleration, the friction force must be zero? Wait, no—wait, when an object moves at constant speed on a surface, if there's no applied force other than friction? Wait, no, the conveyor belt is exerting a force on the box. Wait, maybe I made a mistake. Wait, the box is moving at constant speed, so acceleration is zero. So the net force is zero. The horizontal forces: the friction force and any other force. But if the box is moving with the conveyor belt at constant speed, then the friction force must be zero? Wait, no—wait, when you have an object on a conveyor belt moving at constant speed, the friction force is zero because there's no relative motion and no acceleration. Wait, let's think again. Newton's second law: $F_{net}=ma$. Since $a = 0$, $F_{net}=0$. The vertical forces: $N = mg$. Horizontal forces: the only horizontal force could be friction. But if $F_{net}=0$, then friction must be zero. Because there's no other horizontal force. So the force of friction is zero. Because the box is moving at constant speed, so no net force, so friction (which would be a force) must be balanced by another force, but there is no other horizontal force. So friction force is zero.

# Answer:
The force of friction is $0\ \text{N}$. Justification: The box moves at constant speed, so acceleration $a = 0$. By Newton’s second law ($F_{net}=ma$), net force is zero. Vertically, normal force $N = mg$. Horizontally, no other applied force, so friction force must be zero (to satisfy $F_{net,x}=0$).

### Question 9b
# Explanation:
## Step1: Identify friction type
To prevent slipping, static friction acts. The maximum static friction is $f_{s,max}=\mu_s N$.
## Step2: Find normal force
Normal force $N = mg$ (vertical equilibrium).
## Step3: Apply Newton’s second law
For the box, the static friction provides the acceleration: $f_{s,max}=ma_{max}$. Substitute $f_{s,max}=\mu_s mg$ into $f_{s,max}=ma_{max}$: $\mu_s mg = ma_{max}$. Cancel $m$: $a_{max}=\mu_s g$.
## Step4: Calculate value
$\mu_s = 0.5$, $g = 9.8\ \text{m/s}^2$. So $a_{max}=0.5\times9.8 = 4.9\ \text{m/s}^2$ (direction same as belt’s acceleration, to the right).

# Answer:
The largest acceleration is $4.9\ \text{m/s}^2$ (to the right). Justification: To prevent slipping, static friction ($f_s\leq\mu_s N$) provides the acceleration. $N = mg$, so $f_{s,max}=\mu_s mg$. By $F_{net}=ma$, $\mu_s mg = ma_{max}\Rightarrow a_{max}=\mu_s g = 0.5\times9.8 = 4.9\ \text{m/s}^2$.

### Question 9c
# Explanation:
## Step1: Identify friction type
When the belt stops, the box slides, so kinetic friction acts: $f_k=\mu_k N$.
## Step2: Find acceleration due to friction
By Newton’s second law: $f_k = ma\Rightarrow \mu_k mg = ma\Rightarrow a = \mu_k g$.
## Step3: Calculate time to stop
Initial speed $v_0 = 6\ \text{m/s}$, final speed $v = 0$, acceleration $a = -\mu_k g$ (deceleration). Use $v = v_0 + at$: $0 = 6 - \mu_k g t$. Solve for $t$: $t=\frac{6}{\mu_k g}$. Substitute $\mu_k = 0.3$, $g = 9.8$: $t=\frac{6}{0.3\times9.8}\approx\frac{6}{2.94}\approx2.04\ \text{s}$.
## Step4: Calculate distance slid
Use $v^2 = v_0^2 + 2ax$: $0 = 6^2 + 2(-\mu_k g)x\Rightarrow x=\frac{6^2}{2\mu_k g}=\frac{36}{2\times0.3\times9.8}=\frac{36}{5.88}\approx6.12\ \text{m}$.

# Answer:
Time to stop: $\approx2.04\ \text{s}$; Distance slid: $\approx6.12\ \text{m}$. Justification: Kinetic friction ($f_k=\mu_k mg$) causes deceleration $a = -\mu_k g$. Using kinematics: $v = v_0 + at$ gives $t=\frac{v_0}{\mu_k g}$, and $v^2 = v_0^2 + 2ax$ gives $x=\frac{v_0^2}{2\mu_k g}$. Substituting $v_0 = 6\ \text{m/s}$, $\mu_k = 0.3$, $g = 9.8\ \text{m/s}^2$ yields the values.

### Question 10a
# Explanation:
## Step1: Analyze forces on Box A
For Box A, the static friction from Box B provides its acceleration: $f_{s,max}=\mu_s N_A = m_A a_{max}$, where $N_A = m_A g$ (normal force on A, equal to its weight).
## Step2: Analyze forces on the system (A + B)
Total mass $M = m_A + m_B$. The applied force $F$ causes acceleration $a_{max}$: $F = M a_{max}$.
## Step3: Substitute $a_{max}$ from Box A’s equation
From Box A: $a_{max}=\mu_s g$ (since $f_{s,max}=\mu_s m_A g = m_A a_{max}\Rightarrow a_{max}=\mu_s g$).
## Step4: Calculate F
Substitute $a_{max}$ into $F = (m_A + m_B)\mu_s g$. $m_A = 10\ \text{kg}$, $m_B = 8\ \text{kg}$, $\mu_s = 0.7$, $g = 9.8$: $F=(10 + 8)\times0.7\times9.8 = 18\times6.86 = 123.48\ \text{N}$.

# Answer:
Maximum $F$ is $123.48\ \text{N}$. Work: For Box A, $f_{s,max}=\mu_s m_A g = m_A a_{max}\Rightarrow a_{max}=\mu_s g$. For the system, $F=(m_A + m_B)a_{max}=(m_A + m_B)\mu_s g=(10 + 8)\times0.7\times9.8 = 123.48\ \text{N}$.

### Question 10b
# Explanation:
## Step1: Analyze Box A’s forces
Static friction provides acceleration: $f_{s,max}=\mu_s m_A g = m_A a_{max}$.
## Step2: Solve for $a_{max}$
Cancel $m_A$: $a_{max}=\mu_s g = 0.7\times9.8 = 6.86\ \text{m/s}^2$.

# Answer:
Maximum acceleration is $6.86\ \text{m/s}^2$. Justification: For Box A, static friction ($f_{s,max}=\mu_s m_A g$) causes acceleration $a_{max}$. By $f_{s,max}=m_A a_{max}$, $a_{max}=\mu_s g = 6.86\ \text{m/s}^2$.

### Question 10c
# Explanation:
Original $F=(m_A + m_B)\mu_s g$. New $m_A'$ (greater than original $m_A$). So new $F'=(m_A' + m_B)\mu_s g$. Since $m_A' > m_A$, $F' > F$ (because $(m_A' + m_B) > (m_A + m_B)$ and other terms are constant).

# Answer:
The new maximum $F$ is greater than the answer in part (a). Justification: $F=(m_A + m_B)\mu_s g$. With $m_A'$ (new $m_A$) larger, $(m_A' + m_B) > (m_A + m_B)$, so $F'$ (new $F$) increases (as $\mu_s$ and $g$ are constant).

### Question 10d
# Explanation:
From part (b), $a_{max}=\mu_s g$. This depends only on $\mu_s$ and $g$, not on $m_A$ (since $a_{max}=\mu_s g$ after canceling $m_A$ in $f_{s,max}=m_A a_{max}$). So changing $m_A$ (to a greater mass) does not affect $a_{max}$.

# Answer:
The maximum acceleration remains the same as in part (b). Justification: $a_{max}=\mu_s g$ (derived from $f_{s,max}=\mu_s m_A g = m_A a_{max}$, where $m_A$ cancels). Thus, $a_{max}$ depends only on $\mu_s$ and $g$, not on $m_A$’s mass.

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Question 9a

Step1: Analyze motion state

Step2: Identify friction type

Step1: Identify friction type

Step2: Find normal force

Step3: Apply Newton’s second law

Step4: Calculate value

Step1: Identify friction type

Step2: Find acceleration due to friction

Step3: Calculate time to stop

Step4: Calculate distance slid

Answer:

Question 9b