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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 4.2 The ketchup slide bir In this example the acceleration is caused by a friction force. A counter attendant in a diner shoves a ketchup bottle with mass 0.20 kg along a smooth, level lunch counter. The bottle leaves her hand with an initial velocity of 2.8 m/s. As it slides, it slows down because of the horizontal friction force exerted on it by the countertop. The bottle slides a distance of 1.0 m before coming to rest. What are the magnitude and direction of the friction force acting on it? SOLUTION SET UP Figure 4.14 shows our diagrams: one for the bottle's motion and one for the forces on the bottle. We place the origin at the point where the waitress releases the bottle and point the +x axis in the direc- tion the bottle moves. Because the bottle has no vertical acceleration, we know that the vertical forces on it (the weight and the upward normal force ) must sum to zero. The only horizontal force on the bottle is the friction force f. Because the bottle is slowing down (a, is negative in our coordinate system), we know that this force points in the x direction. We draw one diagram for the bottle's motion and one showing the forces on the bottle. stroncan m= 0.20 kg O Yo = 2.8 m/s |v=0 -1.0 m A FIGURE 4.14 Our sketch for this problem. -X 77777777777 Bo SOLVE We need to find the relationship between u, and a,; to do this, we can use the constant-acceleration equation v voz = 2ax(x - xo) (Equation 2.11). The initial x component of velocity is Vox = 2.8 m/s, and the final value is ux = 0. Also, xo = 0 and x = 1.0 m. Once a, is known, we can use the component form of Newton's second law, Equation 4.8 (EF, = max), to find the magnitude f of the friction force. Because fis the only horizontal component of force, Fx = -f. Using the numerical values in Equation 2.11, we first find the bottle's acceleration: Video Tutor Solution v² - voz = 2a,(x - xo), 0 (2.8 m/s)² = 2a,(1.0m - 0), ax = -3.9 m/s². We can now find the net force acting on the bottle: ΣFx = -f= max, -f= (0.20 kg) (-3.9 m/s²) -0.78 kg m/s² = -0.78 N. Since the friction force is the only horizontal force acting on the bottle, this answer gives us its magnitude (0.78 N) and also the fact that it acts in the -x direction (which we already knew). . CONTINUED
Transcribed Image Text:EXAMPLE 4.2 The ketchup slide bir In this example the acceleration is caused by a friction force. A counter attendant in a diner shoves a ketchup bottle with mass 0.20 kg along a smooth, level lunch counter. The bottle leaves her hand with an initial velocity of 2.8 m/s. As it slides, it slows down because of the horizontal friction force exerted on it by the countertop. The bottle slides a distance of 1.0 m before coming to rest. What are the magnitude and direction of the friction force acting on it? SOLUTION SET UP Figure 4.14 shows our diagrams: one for the bottle's motion and one for the forces on the bottle. We place the origin at the point where the waitress releases the bottle and point the +x axis in the direc- tion the bottle moves. Because the bottle has no vertical acceleration, we know that the vertical forces on it (the weight and the upward normal force ) must sum to zero. The only horizontal force on the bottle is the friction force f. Because the bottle is slowing down (a, is negative in our coordinate system), we know that this force points in the x direction. We draw one diagram for the bottle's motion and one showing the forces on the bottle. stroncan m= 0.20 kg O Yo = 2.8 m/s |v=0 -1.0 m A FIGURE 4.14 Our sketch for this problem. -X 77777777777 Bo SOLVE We need to find the relationship between u, and a,; to do this, we can use the constant-acceleration equation v voz = 2ax(x - xo) (Equation 2.11). The initial x component of velocity is Vox = 2.8 m/s, and the final value is ux = 0. Also, xo = 0 and x = 1.0 m. Once a, is known, we can use the component form of Newton's second law, Equation 4.8 (EF, = max), to find the magnitude f of the friction force. Because fis the only horizontal component of force, Fx = -f. Using the numerical values in Equation 2.11, we first find the bottle's acceleration: Video Tutor Solution v² - voz = 2a,(x - xo), 0 (2.8 m/s)² = 2a,(1.0m - 0), ax = -3.9 m/s². We can now find the net force acting on the bottle: ΣFx = -f= max, -f= (0.20 kg) (-3.9 m/s²) -0.78 kg m/s² = -0.78 N. Since the friction force is the only horizontal force acting on the bottle, this answer gives us its magnitude (0.78 N) and also the fact that it acts in the -x direction (which we already knew). . CONTINUED
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.
Transcribed Image Text: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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