A steel ball (mass = 2.60 kg), is thrown vertically down with initial velocity,

Question

A steel ball (mass = 2.60 kg), is thrown vertically down with initial velocity, and falls a vertical distance of 55 cm before striking a vertical spring. The ball continues to travel downward an additional 15 cm, while compressing the spring, before momentarily coming to rest. The spring constant, K = 1750 N/m.

Using conservation of mechanical energy concepts and ignoring air resistance, find the initial velocity imparted on the steel ball.

After momentarily coming to rest, the spring ejects the ball back upward. What is the maximum height above the platform (i.e., spring at equilibrium) the ball will achieve if air drag is neglected. Note: max height is that height where ball momentarily comes to rest again.

Explanation / Answer

Part 1: energy in =energy out PE+KE= Work done by spring mgh+.5mv^2= .5kx^2 2.60(9.8)(.55+.15) + .5(2.60)v^2 = .5(1750)(.15^2) Solve and you should get v=1.193m/s part 2: energy in= energy out Work done by spring= PE .5(1750)(.15^2)=2.60(9.8)h h=.773m You then need to subtract .773(total height)- .15(height of platform) = .623m(height above platform)

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