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Unformatted text preview: AP Physics Problem Set 7  Fall 200 7 Mr. Shapiro Cha. 7 #12,13,18,29,30,32,33,35,36,42,44,50,62,70,71 due Mon. 10/ 1 12. (a) From Eq. 76, F = W/x = 3.00 N (this is the slope of the graph). (b) Eq. 710 yields K = K i + W = 3.00 J + 6.00 J = 9.00 J. 13. (a) The forces are constant, so the work done by any one of them is given by W F d = ⋅ & & , where & d is the displacement. Force & F 1 is in the direction of the displacement, so 1 1 1 cos (5.00 N)(3.00 m)cos 0 15.0 J. W F d φ = = ° = Force & F 2 makes an angle of 120° with the displacement, so 2 2 2 cos (9.00 N) (3.00 m)cos120 13.5 J. W F d φ = = ° = − Force & F 3 is perpendicular to the displacement, so W 3 = F 3 d cos φ 3 = 0 since cos 90° = 0. The net work done by the three forces is 1 2 3 15.0 J 13.5 J 1.50 J. W W W W = + + = − + = + (b) If no other forces do work on the box, its kinetic energy increases by 1.50 J during the displacement. 18. (a) Using notation common to many vector capable calculators, we have (from Eq. 7 8) W = dot([20.0,0] + [0, − (3.00)(9.8)], [0.500 ∠ 30.0º]) = + 1.31 J. (b) Eq. 710 (along with Eq. 71) then leads to v = 2(1.31 J) / (3.00 kg) = 0.935 m/s. 29. (a) As the body moves along the x axis from x i = 3.0 m to x f = 4.0 m the work done by the force is 2 2 2 2 6 3( ) 3 (4.0 3.0 ) 21 J. f f i i x x x f i x x W F dx x dx x x = = − = − − = − − = − ³ ³ According to the workkinetic energy theorem, this gives the change in the kinetic energy: W K m v v f i = = − ∆ 1 2 2 2 d i where v i is the initial velocity (at x i ) and v f is the final velocity (at x f ). The theorem yields 2 2 2 2( 21) (8.0) 6.6 m/s. 2.0 f i W v v m − = + = + = (b) The velocity of the particle is v f = 5.0 m/s when it is at x = x f . The workkinetic energy theorem is used to solve for x f . The net work done on the particle is ( ) 2 2 3 f i W x x = − − , so the theorem leads to − − = − 3 1 2 2 2 2 2 x x m v v f i f i d i d i ....
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