# chap07 - 7.1 From Eq(7.2 MJ 3.45 J 10 3.45 m(440 s m(9.80...

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Unformatted text preview: 7.1: From Eq. (7.2), . MJ 3.45 J 10 3.45 m) (440 ) s m (9.80 kg) 800 ( 6 2 = × = = mgy 7.2: a) For constant speed, the net force is zero, so the required force is the sack’s weight, N. 49 ) s m kg)(9.80 00 . 5 ( 2 = b) The lifting force acts in the same direction as the sack’s motion, so the work is equal to the weight times the distance, J; 735 m) (15.0 N) 00 . 49 ( = this work becomes potential energy. Note that the result is independent of the speed, and that an extra figure was kept in part (b) to avoid roundoff error. 7.3: In Eq. (7.7), taking 1 = K (as in Example 6.4) and . , other 1 2 2 W U K U + = = Friction does negative work , fy- so ; 2 fy mgy K- = solving for the speed , 2 v . s m 55 . 7 kg) 200 ( m) (3.00 N) 60 ) s m (9.80 kg) 200 (( 2 ) ( 2 2 2 =- =- = m y f mg v 7.4: a) The rope makes an angle of ( 29 ° = 30 arcsin m 6.0 m . 3 with the vertical. The needed horizontal force is then N, 679 30 tan ) s m (9.80 kg) 120 ( tan 2 = ° = θ w or N 10 8 . 6 2 × to two figures. b) In moving the bag, the rope does no work, so the worker does an amount of work equal to the change in potential energy, J. 10 0.95 ) 30 cos (1 m) (6.0 ) s m (9.80 kg) 120 ( 3 2 × = °- Note that this is not the product of the result of part (a) and the horizontal displacement; the force needed to keep the bag in equilibrium varies as the angle is changed. 7.5: a) In the absence of air resistance, Eq. (7.5) is applicable. With m, . 22 2 1 =- y y solving for 2 v gives s. m . 24 m) . 22 )( s m 80 . 9 ( 2 s) m . 12 ( ) ( 2 2 2 1 2 2 1 2 = + =- + = y y g v v b) The result of part (a), and any application of Eq. (7.5), depends only on the magnitude of the velocities, not the directions, so the speed is again s. m . 24 c) The ball thrown upward would be in the air for a longer time and would be slowed more by air resistance. 7.6: a) (Denote the top of the ramp as point 2.) In Eq. (7.7), J, 5 . 87 m) (2.5 N) 35 ( , other 2- = ×- = = W K and taking 1 = U and s, m 25 . 6 J, 147 ) 30 sin m (2.5 ) s m (9.80 kg) 12 ( kg 12 J) 5 . 87 J 147 ( 2 1 2 2 2 = = = ° = = + v mgy U or s m 3 . 6 to two figures. Or, the work done by friction and the change in potential energy are both proportional to the distance the crate moves up the ramp, and so the initial speed is proportional to the square root of the distance up the ramp; . s m 25 . 6 s) m . 5 ( m 1.6 m 2.5 = b) In part a), we calculated other W and 2 U . Using Eq. (7.7), J 491.5 J 147 J 5 . 87 s) m (11.0 kg) 12 ( 2 2 1 2 =-- = K . s m 05 . 9 kg) 12 ( J) 5 . 491 ( 2 2 2 2 = = = m K v 7.7: As in Example 7.7, J, 94 , 2 2 = = U K and . 3 = U The work done by friction is J, 56 m) (1.6 N) 35 (- =- and so J, 38 3 = K and . s m 5 . 2 Kg 12 J) 38 ( 2 3 = = v 7.8: The speed is v and the kinetic energy is 4 K . The work done by friction is proportional to the normal force, and hence the mass, and so each term in Eq. (7.7) is proportional to the total mass of the crate, and the speed at the bottom is the same for any mass. The kinetic energy is proportional to the mass, and for the same speed but four mass....
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## This note was uploaded on 04/08/2008 for the course PHY 205 taught by Professor Nearing during the Fall '07 term at University of Miami.

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chap07 - 7.1 From Eq(7.2 MJ 3.45 J 10 3.45 m(440 s m(9.80...

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