Chapter 06 - 6.1 a 2.40 N(1.5 m 3.60 J b 0.600 N(1.50 m c 3...

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6.1: a) J 60 . 3 ) m 5 . 1 ( ) N 40 . 2 ( = b) J 900 . 0 ) m 50 . 1 )( N 600 . 0 ( - = - c) J 70 . 2 J 720 . 0 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 bucket’s 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 bucket’s 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 0 . 12 )( N 0 . 25 ( = . 6.4: a) The friction force to be overcome is , N 5 . 73 ) s / m 80 . 9 )( kg 0 . 30 )( 25 . 0 ( 2 k k = = = = mg n f μ μ 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 crate’s 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.
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6.5: a) See Exercise 5.37. The needed force is N, 2 . 99 30 sin ) 25 . 0 ( 30 cos ) s / m 80 . 9 )( kg 30 )( 25 . 0 ( sin cos 2 k k = ° - ° = - = φ μ φ μ mg F keeping extra figures. b) J 5 . 386 30 cos ) m 50 . 4 )( N 2 . 99 ( cos = ° = φ Fs , again keeping an extra figure. c) The normal force is φ 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 . 0 )( 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 0 . 15 cos ) m 300 )( N 180 ( cos 4 × = ° = φ Fs 6.7: , J 10 62 . 2 14 cos ) m 10 75 . 0 )( N 10 80 . 1 ( 2 cos 2 9 3 6 × = ° × × = φ Fs or J 10 2.6 9 × to two places. 6.8: The work you do is: ) ˆ ) m 0 . 3 ( ˆ ) m 0 . 9 (( ) ˆ ) N 40 ( ˆ ) N 30 (( j i j i s F - - - = ) m 0 . 3 )( N 40 ( ) m 0 . 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 = , 0 = 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.
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6.10: a) From Eq. (6.6), J. 10 54 . 1 h / km s / m 6 . 3 1 ) km/h 0 . 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. 6.11: For the T-Rex, J 10 32 . 4 ) ) hr / km 4 )(( kg 7000 ( 3 2 km/hr 6 . 3 s / m 1 2 1 × = = K . The person’s velocity would be m/s 1 11 kg J)/70 10 32 . 4 ( 2 3 . v = × = , or about 40 km/h. 6.12: (a) Estimate: 1m s v / (walking) 2m s v / (running) 70kg m Walking: J 35 ) s / m 1 )( kg 70 ( 2 2 1 2 2 1 = = = mv KE Running: J 140 ) s / m 2 )( kg 70 ( 2 2 1 = = KE (b) Estimate: s / m 30 s / ft 88 mph 60 = v kg 2000 m J 10 9 ) s / m 30 )( kg 2000 ( 5 2 2 1 × = = KE (c) mgh W KE = = gravity Estimate m 2 h J 20 ) m 2 )( s / m 8 . 9 )( kg 1 ( 2 = KE 6.13: Let point 1 be at the bottom of the incline and let point 2 be at the skier.
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