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Chapter6

# Chapter6 - CHAPTER 6 WORK AND ENERGY CONCEPTUAL QUESTIONS 1...

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CHAPTER 6 WORK AND ENERGY CONCEPTUAL QUESTIONS ____________________________________________________________________________________________ 1. REASONING AND SOLUTION The work done by F 1 in moving the box through a displacement s is 1 1 1 ( cos 0 ) W F s F s = ° = . The work done by F 2 is 2 2 ( cos ) . W F s θ = From the drawing, we see that 1 2 cos F F θ ; therefore, the force F 1 does more work. ____________________________________________________________________________________________ 2. REASONING AND SOLUTION The force P acts along the displacement; therefore, it does positive work. Both the normal force F N and the weight m g are perpendicular to the displacement; therefore, they do zero work. The kinetic frictional force f k acts opposite to the direction of the displacement; therefore, it does negative work. ____________________________________________________________________________________________ 3. REASONING AND SOLUTION Work can be positive or negative. Work is positive when the force has a component in the same direction as the direction of the displacement, or equivalently, when the angle θ between the force F and the displacement s is less than 90°. The work is negative when the force has a component in the direction opposite to the displacement; that is, when θ is greater than 90°. Since the force does positive work on a particle that has a displacement pointing in the + x direction, the force must have an x component that points in the + x direction. Furthermore, since the same force does negative work on a particle that has a displacement pointing in the + y direction, the force has a y component that points in the negative y direction. When the x and y components of this force are sketched head to tail, as in the figure at the right, we see that the force must point in the fourth quadrant. F x F y F ____________________________________________________________________________________________ 4. SSM REASONING AND SOLUTION The sailboat moves at constant velocity, and, therefore, has zero acceleration. From Newton's second law, we know that the net external force on the sailboat must be zero. a. There is no work done on the sailboat by a zero net external force. b. Work is done by the individual forces that act on the boat; namely the wind that propels the boat forward and the water that resists the motion of the boat. Since the wind propels the boat forward, it does positive work on the boat. Since the force of the water is a resistive force, it acts opposite to the displacement of the boat, and, therefore, it does negative work.

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21 WORK AND ENERGY Since the total work done on the boat is zero, each force must do an equal amount of work with one quantity being positive, and the other being negative. Note: The answer to part (a) could have been deduced from the work-energy theorem as well. Since the velocity of the boat is constant, the kinetic energy of the boat does not change and the total work done on the boat is zero. ____________________________________________________________________________________________ 5. REASONING AND SOLUTION The speed of the ball decreases; therefore, the ball is subjected to an external resistive force. A resistive force always points opposite to the direction of the displacement of the ball. Therefore, the external force does negative work.
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