Recit%207%2c%20chapter%206

Recit%207%2c%20chapter%206 - PHY2053 RECITATIONS, SPRING...

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PHY2053 RECITATIONS, SPRING 2011 Recitation 7 (Chapter 6) Feb. 1-Mar. 4 * 9. ssm A 2.40 × 10 2 -N force is pulling an 85.0-kg refrigerator across a horizontal surface. The force acts at an angle of 20.0 o above the surface. The coefficient of kinetic friction is 0.200, and the refrigerator moves a distance of 8.00 m. Find (a) the work done by the pulling force, and (b) the work done by the kinetic frictional force. 9. REASONING AND SOLUTION a. According to Equation 6.1, the work done by the applied force is W = Fs cos θ = (2.40 × 10 2 N)(8.00 m) cos 20.0° = 1.80 × 10 3 J b. According to Equation 6.1, the work done by the frictional force is W f = f k s cos , where Note the typo below: it should be μ k k s 2 2 2 = ( sin ) = (0.200) (85.0 kg)(9.80 m/s ) (2.40 10 N)(sin 20.0°) = 1.50 10 N f µ mg F - - × × Therefore, W f = (1.50 × 10 2 N)(8.00 m) cos 180° = - 1.20 × 10 3 J ________________________________________________________________________________ * 11. A husband and wife take turns pulling their child in a wagon along a horizontal sidewalk. Each exerts a constant force and pulls the wagon through the same displacement. They do the same amount of work, but the husband’s pulling force is directed 58 o above the horizontal, and the wife’s pulling force is directed 38 o above the horizontal. The husband pulls with a force whose magnitude is 67 N. What is the magnitude of the pulling force exerted by his wife? REASONING The work done by a constant force of magnitude F that makes an angle θ with a displacement of magnitude s is given by ( 29 cos W F s = (Equation 6.1). The magnitudes and directions of the pulling forces exerted by the husband and wife differ, but the displacement of the wagon is the same in both cases. Therefore, we can express the work W H done by the husband’s pulling force as ( 29 H H H cos W F s = . Similarly, the work done by the wife can be written as ( 29 W W W cos W F s = . We know the directional angles of both forces, the magnitude F H of the husband’s pulling force, and that both forces do the same amount of work ( W W = W H ). SOLUTION Setting the two expressions for work equal to one another, and solving for the magnitude F W of the wife’s pulling force, we find ( 29 W W cos F s ( 29 W H H cos W F s = 1 442 4 43 ( 29 ( 29 H H H W W 67 N cos58 cos or 45 N cos cos38 W F F = = = o o 1 442 4 43 1
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________________________________________________________________________________ 20. An asteroid is moving along a straight line. A force acts along the displacement of the asteroid and slows it down. The asteroid has a mass of 4 4.5 10 kg, × and the force causes its speed to change from 7100 to 5500 m/s. (a)
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This note was uploaded on 09/21/2011 for the course COP 3330 taught by Professor Staff during the Spring '08 term at University of Central Florida.

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Recit%207%2c%20chapter%206 - PHY2053 RECITATIONS, SPRING...

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