A block with mass m =7.0 kg is hung from a vertical spring. When the mass hangs

Question

A block with mass m =7.0 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.28 m. While at this equilibrium position, the mass is then given an initial push downward at v = 4.1 m/s. The block oscillates on the spring without friction. 1. What is the spring constant of the spring? A: 245 N/m 2.What is the oscillation frequency? A; 0.94339 Hz 3. After t = 0.37 s what is the speed of the block? 4. What is the magnitude of the maximum acceleration of the block? 5. At t = 0.37 s what is the magnitude of the net force on the block?

Explanation / Answer

1) The spring constant K is computed about the mass at rest

F = kx = m*g = k*.2

k = m*g/0.2

= 7.0 *9.81/0.2

= 343.35 N/m


2)The frequency of oscillation is

f = sqrt( k/m ) / ( 2* )

= sqrt( 245 /7 ) / ( 2* )

= 0.941 Hz

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