A 9.7-kg watermelon and a 7.0-kg pumpkin are attached to each other via a cord t

Question

A 9.7-kg watermelon and a 7.0-kg pumpkin are attached to each other via a cord that wraps over a pulley, as shown. Friction is negligible everywhere in this system. (a) Find the accelerations of the pumpkin and the watermelon. Specify magnitude and direction. magnitude direction pumpkin m/s2 watermelon m/s2 (b) If the system is released from rest, how far along the incline will the pumpkin travel in 0.30 s? cm (c) What is the speed of the watermelon after 0.20 s? cm/s

pumpkin at an incline of /_ 53 degrees and watermelon at an incline of _ 30 degrees

Explanation / Answer

Let T be the tension in the cord and a be the acceleration (which has the same magnitude for each item) Assume the watermelon moves down (If the acceleration turns out to be negative then it moves up) A free body diagram about the melon we get m*g*sin(53) - T = m*a or T = 8.7*9.8*sin(53) - 8.7*a About the pumpkin we get T - m*g*sin(30) = m*a or T = 7.0*9.8*sin(30) + 7.0*a Equating the equations we get 8.7*9.8*sin(53) - 8.7*a = 7.0*9.8*sin(30) + 7.0*a So (8.7+7.0)*a = 8.7*9.8*sin(53) - 7.0*9.8*sin(30) Or a = ( 8.7*9.8*sin(53) - 7.0*9.8*sin(30))/15.7 = 2.15m/s^2 (Pumpkin up and Watermelon down) b) The pumpkin will travel d = 1/2*a*t^2 = 1/2*2.15*0.30^2 = 9.69x10^-2m = 9.69cm c) v = a*t = 2.15m/s^2*0.20s = 0.430m/s = 43 cm/s

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