The block is con nected by a massless string over a

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. The block is con- nected by a massless string over a massless, fric- tionless pulley to a block of mass m 2 = 8 . 60 kg. When the system is released from rest, m 2 accel- erates downward at 1 . 40 m/s 2 . m 1 m 2 θ (a) Draw neat, careful diagrams for each mass, showing all the forces acting on each. Include coordinate axes for each diagram. (8 pts) Physical picture style diagrams are shown, but free-body diagrams are also accept- able. m 1 g F N F T m 2 g F T m 1 m 2 +x +y +x +y (b) For each object, write down Newton’s Second Law equations for each of your chosen axes. (9 points) F T - m 1 g sin θ = m 1 a m 2 g - F T = m 2 a (1) F N - m 1 g cos θ = 0 0 = m 2 (0) (2) (c) Find the tension in the string and the mass m 1 . (11 points) The x (vertical) equation for m 2 can be used to solve for F T : F T = m 2 g - m 2 a = m 2 ( g - a ) = (8 . 60 kg)(9 . 81 m/s 2 - 1 . 40 m/s 2 ) = 72 . 326 N 72 . 3 N . Now use the x equation for m 1 , and substitute the previous result: m 1 g sin θ + m 1 a = F T = m 1 ( g sin θ + a ) = m 2 ( g - a ) m 1 = m 2 ( g - a ) g sin θ + a = (8 . 60 kg)(9 . 81 m/s 2 - 1 . 40 m/s 2 ) (9 . 81 m/s 2 ) sin(35 . 0 ) + 1 . 40 m/s 2 = 10 . 2929 kg 10 . 3 kg . Copyright c 2017 University of Georgia. 4
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Physics 1111 Exam #2 2:30-B Solutions 13 October 2017 Last Name: IV: Sledding (18 points) After an exciting sled ride, Calvin and Hobbes are heading home. Calvin pulls the sled with a force of 15 . 2 N at an angle of 22 . 0 above the horizontal, while Hobbes pushes horizontally from the rear with a force of 20 . 4 N. On their way home, they cross a flat field 39 . 0 m across. The sled has a mass of 9 . 10 kg (a) How much total work do Calvin and Hobbes do on the sled as they move it across the field? (9 points) W C = F C x ) cos θ C = (15 . 2 N)(39 . 0 m) cos(22 . 0 ) = 549 . 635 J 550 . J . W H = F H x ) = (20 . 4 N)(39 . 0 m) = 795 . 600 J 796 J . W tot = W C + W H = 1345 . 235 J 1 . 35 × 10 3 J . Alternatively, we can calculate the net force along the horizontal direction and calculate the work that way. F x = F C cos θ C + F H = 34 . 493 N; W tot = F x x ) (b) Say that Calvin and Hobbes take a break to rest just before starting across the field. If we assume there is no friction between the sled and the snow, what speed would the sled have after crossing the field? Does this seem reasonable, or should we reconsider our assumptions? (9 points) W tot = Δ K = 1 2 mv 2 f - 1 2 m (0) 2 v f = r 2 W tot m = s 2(1345 . 235 J) 9 . 10 kg = 17 . 1946 m/s 17 . 2 m/s .
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