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A 17.5 kg block is dragged over a rough, horizontal surface by a 75 N force acting at 21° above the horizontal. The block is displaced 5.7 m, and the coefficient of kinetic friction is 0.200. (Enter your answers in J.) Hint (a) Find the work done on the block by the 75 N force. (b) Find the work done on the block by the normal force. (c) Find the work done on the block by the gravitational force. (d) What is the increase in internal energy of the block-surface system due to friction? (e) Find the total change in the block's kinetic energy.

A 17.5 kg block is dragged over a rough, horizontal surface by a 75 N force acting at 21° above the horizontal. The block is displaced 5.7 m, and the coefficient of kinetic friction is 0.200. (Enter your answers in J.) Hint (a) Find the work done on the block by the 75 N force. (b) Find the work done on the block by the normal force. (c) Find the work done on the block by the gravitational force. (d) What is the increase in internal energy of the block-surface system due to friction? (e) Find the total change in the block's kinetic energy.

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Categorize The block is pulled by a force and the surface is rough, so the block and the surface are modeled as ---Select-- |system with a nonconservative force. A Block Pulled on a Rough Surface A 5.0 kg block initially at rest is pulled to the right along a horizontal surface by a constant horizontal force of Analyze Figure (a) illustrates this situation. Neither the normal force nor the gravitational force does work on the system because their points of application are displaced horizontally. 21 N. Find the work done on the system by the applied force (Use the following as necessary: F and Ax.): (a) A block pulled to the right on a rough SWother forces- W. surface by a constant horizontal force. Apply the particle in equilibrium model to the block in the vertical direction (Use the following as necessary: F, Ax, m, and g.): ▼ mg Efy =0-n - mg = 0 → n = (b) The applied force is at an angle o to the Find the magnitude of the friction force (in N): horizontal. Substitute the energies into the equation shown and solve for the final speed of the block (Use the following as necessary: F, Ax and f): FAx = AK + AEint - Gmv? - 0) + ,ax mg V -1,Ax + m (For the following, when entering a mathematical expression, do not substitute numerical values; use variables only.) (a) Find the speed of the block after it has moved 2.5 m if the surfaces in contact have a coefficient of kinetic friction of 0.16. Substitute numerical values (Enter your answer in m/s.): ]m/s SOLUTION Finalize As expected, this value is ---Select--- v that found in the case of the block sliding on a Conceptualize This example is similar to the example "A Block Pulled on a Frictionless Surface" in Chapter 7, but modified so that the surface is no longer frictionless. frictionless surface. The difference in kinetic energies between the block sliding on a frictionless surface and the block in this example is equal to the increase in internal energy of the block-surface system in this example. The rough surface applies a friction force on the block opposite to the applied force. As a result, we expect the speed to be ---Select--- v that found in "A Block Pulled on a Frictionless Surface." (b) Suppose the force F is applied at an angle 0 as shown in figure (b). At what angle should the force be applied to achieve the largest possible speed after the block has moved 2.5 m to the right?
Transcribed Image Text:Categorize The block is pulled by a force and the surface is rough, so the block and the surface are modeled as ---Select-- |system with a nonconservative force. A Block Pulled on a Rough Surface A 5.0 kg block initially at rest is pulled to the right along a horizontal surface by a constant horizontal force of Analyze Figure (a) illustrates this situation. Neither the normal force nor the gravitational force does work on the system because their points of application are displaced horizontally. 21 N. Find the work done on the system by the applied force (Use the following as necessary: F and Ax.): (a) A block pulled to the right on a rough SWother forces- W. surface by a constant horizontal force. Apply the particle in equilibrium model to the block in the vertical direction (Use the following as necessary: F, Ax, m, and g.): ▼ mg Efy =0-n - mg = 0 → n = (b) The applied force is at an angle o to the Find the magnitude of the friction force (in N): horizontal. Substitute the energies into the equation shown and solve for the final speed of the block (Use the following as necessary: F, Ax and f): FAx = AK + AEint - Gmv? - 0) + ,ax mg V -1,Ax + m (For the following, when entering a mathematical expression, do not substitute numerical values; use variables only.) (a) Find the speed of the block after it has moved 2.5 m if the surfaces in contact have a coefficient of kinetic friction of 0.16. Substitute numerical values (Enter your answer in m/s.): ]m/s SOLUTION Finalize As expected, this value is ---Select--- v that found in the case of the block sliding on a Conceptualize This example is similar to the example "A Block Pulled on a Frictionless Surface" in Chapter 7, but modified so that the surface is no longer frictionless. frictionless surface. The difference in kinetic energies between the block sliding on a frictionless surface and the block in this example is equal to the increase in internal energy of the block-surface system in this example. The rough surface applies a friction force on the block opposite to the applied force. As a result, we expect the speed to be ---Select--- v that found in "A Block Pulled on a Frictionless Surface." (b) Suppose the force F is applied at an angle 0 as shown in figure (b). At what angle should the force be applied to achieve the largest possible speed after the block has moved 2.5 m to the right?
SOLUTION Conceptualize You might guess that 0 = 0 would give the largest speed because the force would have the largest component possible in the direction parallel to the surface. Think about F applied at an arbitrary nonzero angle, however. Although the horizontal component of the force would be reduced, the vertical component of the force would ---Select--- v the normal force, in turn reducing the force of friction, which suggests that the speed could be maximized by pulling at an angle other than 8 = 0. Categorize As in part (a), we model the block and the surface as a nonisolated system with a ---Select--- 0 = tan-(Hg) = tan-'(0.16) = | v force acting. Finalize Notice that the angle at which the speed of the block is a maximum is indeed not e = 0. When the angle exceeds 9.09°, the horizontal component of the applied force is too small to be compensated by the reduced friction force and the speed of the block begins to ---Select--- v from its maximum Analyze Find the work done by the applied force, noting that d = Ax because the path followed by the block is a straight line (Use the following as necessary: F, Ax, and 0.): value. "other forces = W = Fd cos(0) = EXERCISE (1) A 17.5 kg block is dragged over a rough, horizontal surface by a 75 N force acting at 21° above the horizontal. Apply the particle in equilibrium model to the block in the vertical direction: The block is displaced 5.7 m, and the coefficient of kinetic friction is 0.200. (Enter your answers in J.) SFy =n + F sin(0) - mg = 0 Hint Solve for n (Use the following as necessary: F, Ax, 0, m and g.): (a) Find the work done on the block by the 75 N force. (2) n = (b) Find the work done on the block by the normal force. Use Wother forces - W = AK + AEnt to find the final kinetic energy for this situation: (c) Find the work done on the block by the gravitational force. WE = AK + AEint = (Kf - 0) + fAx → K = WF - fAx Substitute the results in Equations (1) and (2): Kf = FAx cos(0) – HynAx = FAx cos(0) – Hg(mg – F sin(0))Ax (d) What is the increase in internal energy of the block-surface system due to friction? Maximizing the speed is equivalent to maximizing the final kinetic energy. Consequently, differentiate K, with respect to e and set the result equal to zero (Use the following as necessary: Ax, 0, and g.): dKf -= -FAx sin(0) - H(0 - F cos(0))Ax = 0 de (e) Find the total change in the block's kinetic energy. -sin(0) + Hy cos(0) = 0 = H Need Help? Read It Evaluate e (in degrees) for uk = 0.16:
Transcribed Image Text:SOLUTION Conceptualize You might guess that 0 = 0 would give the largest speed because the force would have the largest component possible in the direction parallel to the surface. Think about F applied at an arbitrary nonzero angle, however. Although the horizontal component of the force would be reduced, the vertical component of the force would ---Select--- v the normal force, in turn reducing the force of friction, which suggests that the speed could be maximized by pulling at an angle other than 8 = 0. Categorize As in part (a), we model the block and the surface as a nonisolated system with a ---Select--- 0 = tan-(Hg) = tan-'(0.16) = | v force acting. Finalize Notice that the angle at which the speed of the block is a maximum is indeed not e = 0. When the angle exceeds 9.09°, the horizontal component of the applied force is too small to be compensated by the reduced friction force and the speed of the block begins to ---Select--- v from its maximum Analyze Find the work done by the applied force, noting that d = Ax because the path followed by the block is a straight line (Use the following as necessary: F, Ax, and 0.): value. "other forces = W = Fd cos(0) = EXERCISE (1) A 17.5 kg block is dragged over a rough, horizontal surface by a 75 N force acting at 21° above the horizontal. Apply the particle in equilibrium model to the block in the vertical direction: The block is displaced 5.7 m, and the coefficient of kinetic friction is 0.200. (Enter your answers in J.) SFy =n + F sin(0) - mg = 0 Hint Solve for n (Use the following as necessary: F, Ax, 0, m and g.): (a) Find the work done on the block by the 75 N force. (2) n = (b) Find the work done on the block by the normal force. Use Wother forces - W = AK + AEnt to find the final kinetic energy for this situation: (c) Find the work done on the block by the gravitational force. WE = AK + AEint = (Kf - 0) + fAx → K = WF - fAx Substitute the results in Equations (1) and (2): Kf = FAx cos(0) – HynAx = FAx cos(0) – Hg(mg – F sin(0))Ax (d) What is the increase in internal energy of the block-surface system due to friction? Maximizing the speed is equivalent to maximizing the final kinetic energy. Consequently, differentiate K, with respect to e and set the result equal to zero (Use the following as necessary: Ax, 0, and g.): dKf -= -FAx sin(0) - H(0 - F cos(0))Ax = 0 de (e) Find the total change in the block's kinetic energy. -sin(0) + Hy cos(0) = 0 = H Need Help? Read It Evaluate e (in degrees) for uk = 0.16:
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