Page 102 Practice Problem 4.2: Suppose the counter attendant pushes a 0.35kg bottle with the same initial speed on a different countertop and it travels 1.5m before stopping. What is the magnitude of the friction force from this second counter? Answer: 0.92N.
Page 102 Practice Problem 4.2: Suppose the counter attendant pushes a 0.35kg bottle with the same initial speed on a different countertop and it travels 1.5m before stopping. What is the magnitude of the friction force from this second counter? Answer: 0.92N.
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![**Example 4.12: The Ketchup Slide**
In this example, the acceleration is caused by friction (specifically a force exerted on a counter). A waiter slides a ketchup bottle, with a mass of 0.20 kg, along a smooth, level lunch counter. The bottle leaves his hand with an initial velocity of 2.8 m/s and eventually slows down due to horizontal friction. The bottle slides a distance of 0.10 m before coming to rest. What are the magnitude and direction of the friction force acting on it?
**Setup**
Figure 4.14 illustrates two diagrams: one for the bottle's motion and another showing the forces on the bottle. The initial position, \( x_0 \), is the point where the waiter releases the bottle, and the point \( x \) is the point where the bottle stops. The bottle does not move vertically, so there is no vertical acceleration, and the normal force (\( n \)) and gravitational force (\( mg \)) sum to zero. The only horizontal force acting is the friction force (\( f \)), which acts counter to the direction of motion. Since the bottle is slowing down, its acceleration (\( a \)) is negative.
**Diagram Details**
1. Top Diagram: Shows a ketchup bottle sliding to the right with an initial velocity \( v_0 = 2.8 m/s \), mass \( m = 0.20 kg \), and coming to rest at \( v = 0 \) after traveling 0.10 m.
2. Bottom Diagram: Illustrates the forces on the bottle. The normal force \( n \) acts upward, gravitational force \( mg \) acts downward, and friction force \( f \) acts to the left.
**Solution Steps**
1. **Use Kinematic Equation**: We use the constant-acceleration equation to find acceleration (\( a \)):
\[
v^2 = v_0^2 + 2a(x - x_0)
\]
Plugging in the known values,
\[
0 = (2.8 m/s)^2 + 2a(0.10m)
\]
Solving for \( a \),
\[
a = -39 m/s^2
\]
2. **Apply Newton's Second Law**: Using the horizontal component,
\[
\Sigma F_x = ma](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F28764828-4d1c-44b8-9544-7d79c8b69e6b%2F2b857aae-aa56-4215-b179-cb31042e1546%2Fotrrzsg_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Example 4.12: The Ketchup Slide**
In this example, the acceleration is caused by friction (specifically a force exerted on a counter). A waiter slides a ketchup bottle, with a mass of 0.20 kg, along a smooth, level lunch counter. The bottle leaves his hand with an initial velocity of 2.8 m/s and eventually slows down due to horizontal friction. The bottle slides a distance of 0.10 m before coming to rest. What are the magnitude and direction of the friction force acting on it?
**Setup**
Figure 4.14 illustrates two diagrams: one for the bottle's motion and another showing the forces on the bottle. The initial position, \( x_0 \), is the point where the waiter releases the bottle, and the point \( x \) is the point where the bottle stops. The bottle does not move vertically, so there is no vertical acceleration, and the normal force (\( n \)) and gravitational force (\( mg \)) sum to zero. The only horizontal force acting is the friction force (\( f \)), which acts counter to the direction of motion. Since the bottle is slowing down, its acceleration (\( a \)) is negative.
**Diagram Details**
1. Top Diagram: Shows a ketchup bottle sliding to the right with an initial velocity \( v_0 = 2.8 m/s \), mass \( m = 0.20 kg \), and coming to rest at \( v = 0 \) after traveling 0.10 m.
2. Bottom Diagram: Illustrates the forces on the bottle. The normal force \( n \) acts upward, gravitational force \( mg \) acts downward, and friction force \( f \) acts to the left.
**Solution Steps**
1. **Use Kinematic Equation**: We use the constant-acceleration equation to find acceleration (\( a \)):
\[
v^2 = v_0^2 + 2a(x - x_0)
\]
Plugging in the known values,
\[
0 = (2.8 m/s)^2 + 2a(0.10m)
\]
Solving for \( a \),
\[
a = -39 m/s^2
\]
2. **Apply Newton's Second Law**: Using the horizontal component,
\[
\Sigma F_x = ma
Transcribed Image Text:**Page 102 Practice Problem 4.12:**
Suppose the counter attendant pushes a 0.35 kg bottle with the same initial speed on a different countertop and it travels 1.5 m before stopping. What is the magnitude of the friction force from this second counter?
**Answer: 0.92 N.**
(Note: There are no graphs or diagrams associated with this text.)
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