Problem 4:The puck in the figure above on the left has a mass of 0.120 kg. The distance of the puck fromthe center of rotation is originally 0.4 m, and the puck is sliding with a speed of v₁ = 0.8 m/sec.The string is pulled downward through the hole in the frictionless table and the new distance ofthe puck from the center of rotation is 0.25 m. What is the new speed of the puck?Answer: v= 1.28 m/secVf0Rm KE = mv², KE = 1w², Ug = mgh, U₁ = ¹kx², E = KE + Ug + Us, E₁ = EfL = mrv, 1 =1w

The puck of mass 0.12 kg is initially moving in a 0.4 m radius circle with a linear velocity of 0.8…

Answered: Problem 4: The puck in the figure above…

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