A 5.0 kg block at rest on a tabletop is attached to a horizontal spring having c

Question

A 5.0 kg block at rest on a tabletop is attached to a horizontal spring having constant 19.6 N/m as in Figure P13.12. The spring is initially unstretched. A constant 20 N horizontal force is applied to the object, causing the spring to stretch.

Figure 13.12 --> http://www.webassign.net/sf/p13_12.gif
(a) Determine the speed of the block after it has moved 1.000 m from equilibrium if the surface between the block and tabletop is frictionless. (in m/s)
(b) Answer part (a) if the coefficient of kinetic friction between block and tabletop is 0.200. (in m/s)

Explanation / Answer

   from the theory according to the work energytheorem we get    Wnc = (KE + PEg +PEs)f - (KE + PEg +PEs)i    F. xf = (1 / 2) mvf2 + 0 + (1 / 2) kxf2    the speed of the block will be    vf = [(2 F . xf - kxf2) / m]        ={[2(20.0 N) (1.000 m) - (19.6 N / m) (1.000m)2] / (5.0 kg)}        = 2.019 m / s (b)
for this we can still use energy by simply accounting for another negative force
   vf = {[2 (F -k m g) . xf - kxf2] / m}
      = {[2(20.0-0.2*9.8*5.0) (1.000 m) - (19.6 N / m) (1.000m)2] / (5.0 kg)}
      = 0.4 m/s

   
   F. xf = (1 / 2) mvf2 + 0 + (1 / 2) kxf2    the speed of the block will be    vf = [(2 F . xf - kxf2) / m]        ={[2(20.0 N) (1.000 m) - (19.6 N / m) (1.000m)2] / (5.0 kg)}        = 2.019 m / s (b)
for this we can still use energy by simply accounting for another negative force
   vf = {[2 (F -k m g) . xf - kxf2] / m}
      = {[2(20.0-0.2*9.8*5.0) (1.000 m) - (19.6 N / m) (1.000m)2] / (5.0 kg)}
      = 0.4 m/s

   

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