lass Management Help N5 Begin Date: 221/2018 120100 AM-Due Date: 3/12018 11.59.0

Question

lass Management Help N5 Begin Date: 221/2018 120100 AM-Due Date: 3/12018 11.59.00 PM End Date: 5/52018 11-5900 PM (6%) Problem 14 As datp hes a baseball of"-015kg down onto the top of a vertical spring that has its lower end fixed to a table, compressing the spring a distance of d- 0.18 meters. The gravitational potential energy be zero at the position of the baseball in the compressed spring spring constant of the spring is k-910 N/m. Let the Randomized Variables = 0.15 kg =910 Nm d-018 m 33% Part (a) The ball is then released what is its speed, vmmeters per second just af er the ball leaves the spring? 33% Part (b) What is the maximum height, h, m meters, that the ball reaches above the equilibnum point? 33% Part (c) what is the ball's velocity in meters per second, at half of the ma amum height relative to the equilibrium po Grade Deduc Potenti Submi Attem cotan0 sin acos0 atan acotan0 sinbo detaile Degrees O Radians pe up Hints:i dedaction per hint Hints remaining 3 Feedback: 1 for a jgs deduction The ansmer is not simply half of the balls velocity just after relesse The elastic potential energy of the spring was converted into the gravitacional potential energy at half be macimum beight and the kinetic energy ar that poiat. LUse your reult fhom Part (c) in

Explanation / Answer

a) From energy conservation

1/2 m v2 = 1/2 k x2

v2 = k x2 / m = (910 * 0.182) / (0.15)

speed v = 14.0 m/s

b) v2 - u2 = 2 a s

s = v2 / 2 a = (142) / (2 * 9.8)

maximum height = 10 m

c) v2 - u2 = 2 a s

v2 - 142 = 2 * -9.8 * (10 / 2)

ball's velocity = 9.89 m/s

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