A 0.20 kg mass is hung from a vertical spring of forceconstant 55N/m. When the s

Question

A 0.20 kg mass is hung from a vertical spring of forceconstant 55N/m. When the spring is released from its unstretchedequilibrium position, the mass is allowed to fall. Use law ofconservation of energy to determine: a) speed of mass after it falls 1.5cm b)distance the mass will fall before reversing direction A 0.20 kg mass is hung from a vertical spring of forceconstant 55N/m. When the spring is released from its unstretchedequilibrium position, the mass is allowed to fall. Use law ofconservation of energy to determine: a) speed of mass after it falls 1.5cm b)distance the mass will fall before reversing direction

Explanation / Answer

loss of gravitational PE = gain in KE + gain in elastic PE mgh = (1/2)mv2 + (1/2)kx2 2gh - kx2/m = v2 v = (2gh - kx2/m) plug in g=9.8, x=h=.015, k=55, m=.2 and compute. if it falls H before changing direction, then since velocity is 0at the lowest point, 0 = (2gH - kH2/m) 2gH = kH2/m H = 2mg/k compute this.

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