Use energy methods to solve the following! A 2.00 kg block spring system is sitt
ID: 1294343 • Letter: U
Question
Use energy methods to solve the following! A 2.00 kg block spring system is sitting on an incline with a rough surrface mu k = 0.18). The incline angle is 23.0 degree. The block is initially making contact but is not permanently attached to the spring. The spring has a spring constant of 500.0N/m. If the block -spring is initially compressed to 12.0cm and released from rest, how far from the equilibrium position of the spring will the block travel up the incline before it stops? (Note: when the block moves up the incline and the spring reaches its equilibrium position, the block is no longer in contact with the spring) A ball of mass 3.00kg is thrown straight upward with an initial speed of 20.0m/s. Use energy methods to determine the following. a) the potential energy of the ball at a height of 5.00m, 10.0m and 18.0m. b) the speed of the ball when it reaches a height of 5.00m,10.0m and 18.0m. c) the height of the ball when it has reached a speed of 18.0m/s, 10.0m/s and 5.00m/s. d) the mechanical energy of the ball when it is first thrown upward, the mechanical energy of the ball when it reaches a speed of 10.0mis and the mechanical energy of the ball when it reaches a height of 18.0m.Explanation / Answer
Let S be the distance (along the incline) that block moves from the compressed position
Wf = S u m g * cos 23 where Wf = work done against friction and u the coefficient of friction
Wh = S m g * sin 23 where Wh is the work done lifting the block vertically
W = Wf + Wh = S m g (u cos 23 + sin 23)
W = S * 2 * 9.8 (.166 + .391) = 10.9 S
W = 1/2 k x^2 potential energy of the compressed spring
W = 1/2 * 500 * .12^2 = 3.6 J
S = 3.6 / 10.9 = .33 m total distance moved by block
Note that block travels .33 - .12 = .21 m after leaving spring
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