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Unformatted text preview: 6.1: a) J 60 . 3 ) m 5 . 1 ( ) N 40 . 2 ( = b) J 900 . ) m 50 . 1 )( N 600 . ( = c) J 70 . 2 J 720 . J 60 . 3 = . 6.2: a) Pulling slowly can be taken to mean that the bucket rises at constant speed, so the tension in the rope may be taken to be the buckets weight. In pulling a given length of rope, from Eq. (6.1), J. 6 . 264 ) m 00 . 4 )( s / m 80 . 9 ( ) kg 75 . 6 ( 2 = = = = mgs Fs W b) Gravity is directed opposite to the direction of the buckets motion, so Eq. (6.2) gives the negative of the result of part (a), or J 265 . c) The net work done on the bucket is zero. 6.3: J 300 ) m . 12 )( N . 25 ( = . 6.4: a) The friction force to be overcome is , N 5 . 73 ) s / m 80 . 9 )( kg . 30 )( 25 . ( 2 k k = = = = mg n f or 74 N to two figures. b) From Eq. (6.1), J 331 ) m 5 . 4 )( N 5 . 73 ( = = Fs . The work is positive, since the worker is pushing in the same direction as the crates motion. c) Since f and s are oppositely directed, Eq. (6.2) gives J. 331 ) m 5 . 4 )( N 5 . 73 ( = = fs d) Both the normal force and gravity act perpendicular to the direction of motion, so neither force does work. e) The net work done is zero. 6.5: a) See Exercise 5.37. The needed force is N, 2 . 99 30 sin ) 25 . ( 30 cos ) s / m 80 . 9 )( kg 30 )( 25 . ( sin cos 2 k k =  = = mg F keeping extra figures. b) J 5 . 386 30 cos ) m 50 . 4 )( N 2 . 99 ( cos = = Fs , again keeping an extra figure. c) The normal force is sin F mg + , and so the work done by friction is J 5 . 386 ) 30 sin ) N 2 . 99 ( ) s / m 80 . 9 )( kg 30 )(( 25 . )( m 50 . 4 ( 2 = + . d) Both the normal force and gravity act perpendicular to the direction of motion, so neither force does work. e) The net work done is zero. 6.6: From Eq. (6.2), J. 10 22 . 5 . 15 cos ) m 300 )( N 180 ( cos 4 = = Fs 6.7: , J 10 62 . 2 14 cos ) m 10 75 . )( N 10 80 . 1 ( 2 cos 2 9 3 6 = = Fs or J 10 2.6 9 to two places. 6.8: The work you do is: ) ) m . 3 ( ) m . 9 (( ) ) N 40 ( ) N 30 (( j i j i s F  = ) m . 3 )( N 40 ( ) m . 9 )( N 30 ( + = J 150 m N 120 m N 270 = +  = 6.9: a) (i) Tension force is always perpendicular to the displacement and does no work. (ii) Work done by gravity is ). ( 1 2 y y mg When 2 1 y y = , = mg W . b) (i) Tension does no work. (ii) Let l be the length of the string. J 1 . 25 ) 2 ( ) ( 1 2 = = = l mg y y mg W mg The displacement is upward and the gravity force is downward, so it does negative work. 6.10: a) From Eq. (6.6), J. 10 54 . 1 h / km s / m 6 . 3 1 ) km/h . 50 ( ) kg 1600 ( 2 1 5 2 = = K b) Equation (6.5) gives the explicit dependence of kinetic energy on speed; doubling the speed of any object increases the kinetic energy by a factor of four....
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This homework help was uploaded on 04/08/2008 for the course PHY 205 taught by Professor Nearing during the Fall '07 term at University of Miami.
 Fall '07
 Nearing
 Physics

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