A block of mass m = 0.785 kg is fastened to an unstrained horizontal spring whos

Question

A block of mass m = 0.785 kg is fastened to an unstrained horizontal spring whose spring constant is k = 87.4 N/m. The block is given a displacement of +0.146 m, where the + sign indicates that the displacement is along the +x axis, and then released from rest. (a) What is the force (with sign) that the spring exerts on the block just before the block is released? (b) Find the angular frequency of the resulting oscillatory motion. (c) What is the maximum speed of the block? (d) Determine the magnitude of the maximum acceleration of the block.

Explanation / Answer

Given Mass of block m = 0.785 kg Spring constant k = 87.4 N / m Distance travelled by the block  x = 0.146 m a) The force that the spring exerts on the block is                 F  = - k x                     = - ( 87.4 N / m ) ( 0.146 m )                     = - 12.7604 N Here negative sign represents restoring force. ___________________________________________ b) Angular frequency of the resulting oscillatory motion                 = ( k / m ) 1/2                     = [ ( 87.4 N /m ) / ( 0.785 kg ) ] 1/2                       = 10.55 rad /s _____________________________________________ c) According to law of conservation of energy                 P .E  = K .E      ( 1 /2 ) kx 2   = ( 1 /2 ) m v 2  Maximum speed                   v  =   ( k x2  / m ) 1/2                         = ( k / m ) 1/2  x                       = ( 10.55 rad / sec ) ( 0.146 m )                        =  1.54 m/s __________________________________________ d)  Magnitude of the maximum acceleration of the block.                  a   = 2x                       = ( 10.55 rad /s )2 ( 0.146 m )                       = 16.25 m /s2

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