Fall 2018 Exam 2 D.pdf

# W tot δ k 1 2 mv 2 f 1 2 mv 2 i plugging in numbers

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W tot = Δ K = 1 2 mv 2 f - 1 2 mv 2 i Plugging in numbers gives W tot = 41 . 3280 J 41 . 3 J . (b) Find the magnitude of the force Letty exerts on the mat. (11 points) As usual, a picture is useful. From this, we can see that only friction and Letty’s force will do work on Peter. We have the total work done on Peter, and we can find the work done by friction, which will allow us to find the work done by Letty. mg F N f k +x +y F L W f = f k d cos(180 ) = - 53 . 200 J Using the total work, we have W tot = W L + W f = W L = W tot - W f = W tot + | W f | = 41 . 3280 J - ( - 53 . 200 J) = 94 . 5280 J Note that since friction does negative work, the work Letty does is greater than the total work. Finally, we use the definition of work by a constant force: W L = F L d cos θ = F L = W L d cos θ = W tot + | W f | d cos θ = 27 . 9188 N 27 . 9 N . Copyright c 2018 University of Georgia. 6

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Physics 1111 Exam #2 2:30-D Solutions 12 October 2018 Last Name: V: Tetherball (18 points) A tetherball spins around its pole, as shown in the picture. At a particular linear speed v , the rope is found to make a θ = 37 . 0 angle with the pole, and the ball makes a hori- zontal circle around the pole of radius R = 0 . 980 m. The tetherball has a mass of m = 0 . 860 kg. θ R mg F T x y (a) What is the tension in the rope? (9 points) The acceleration will be horizontal and radially in- ward, so choose the x axis to be horizontal and radially inward, and the y axis to be vertical. Now we can set up Newton’s second law along each axis. x : F T sin θ = ma r = mv 2 R y : F T cos θ - mg = ma y = 0 Solving the y equation gives F T = mg cos θ = (0 . 860 kg)(9 . 81 m/s 2 ) cos 37 . 0 = 10 . 5638 N 10 . 6 N . (b) What is the linear speed v of the ball? (9 points) Now we can use the x equation from above to find the linear speed. Note that we can use the result for tension from part (a) to simplify. F T sin θ = mv 2 R = v 2 = F T R sin θ m = mg cos θ R sin θ m = gR tan θ v = p gR tan θ = q (9 . 81 m/s 2 )(0 . 980 m) tan 37 . 0 = 2 . 69156 m/s 2 . 69 m/s . Copyright c 2018 University of Georgia. 7
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