A spring k=200N/m is suspended with its upper end supportedfrom a ceiling . With

Question

A spring k=200N/m is suspended with its upper end supportedfrom a ceiling . With the spring hanging in its equilibriumconfiguration, an object mass=2.0Kg is attached to the lower endand released from rest.Whatis the speed of the object after it has fallen4.0cm? A spring k=200N/m is suspended with its upper end supportedfrom a ceiling . With the spring hanging in its equilibriumconfiguration, an object mass=2.0Kg is attached to the lower endand released from rest.Whatis the speed of the object after it has fallen4.0cm?

Explanation / Answer

initial grav potential energy = final springpotential energy + final kinetic energy .    m g h   = (1/2) kx2   + (1/2) m v2 .     2.0 * 9.80 * 0.04   = (1/2)* 200 * 0.042   + (1/2) * 2.0 *v2 .      0 .784   =     0.16   + v2 .        v2 = 0.624           v =   0.790 m/s is the speed of theobject .        v2 = 0.624           v =   0.790 m/s is the speed of theobject

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