A 2kg block is attached to a spring of force constant 500N/m.The block is pulled

Question

A 2kg block is attached to a spring of force constant 500N/m.The block is pulled 5cm to the right of equilibrium and is thenreleased from rest. Find the speed of the block as it passesthrough equilibrium if: a.) the horizontal surface is frictionless b.)the coefficient of friction between the block and thesurface is .350 Please show work for full points A 2kg block is attached to a spring of force constant 500N/m.The block is pulled 5cm to the right of equilibrium and is thenreleased from rest. Find the speed of the block as it passesthrough equilibrium if: a.) the horizontal surface is frictionless b.)the coefficient of friction between the block and thesurface is .350 Please show work for full points

Explanation / Answer

           Given that the mass of the block is m = 2 kg             Thespring constant is K = 500 N/m            The displancement of spring is x = 5 cm = 5*10-2m       ------------------------------------------------------------------------- (a) If surface is frictional less then             From the conservation of energy                       (1/2)Kx2 = (1/2)mV2                                 Kx2 = mV2                                    V = ( kx2 / m )1/2                                      = ---------- m/s    (b) If the surface has frictionalcoefficent = 0.350                Fromthe conservation of energy                       (1/2)Kx2 - mg*x =(1/2)mV2                                 Kx2 - mg*x =mV2                                    V = ( kx2 / m   -g*x)1/2                                      = ---------- m/s                                                                    = ---------- m/s    (b) If the surface has frictionalcoefficent = 0.350                Fromthe conservation of energy                       (1/2)Kx2 - mg*x =(1/2)mV2                                 Kx2 - mg*x =mV2                                    V = ( kx2 / m   -g*x)1/2                                      = ---------- m/s                                                     (1/2)Kx2 - mg*x =(1/2)mV2                                 Kx2 - mg*x =mV2                                    V = ( kx2 / m   -g*x)1/2                                      = ---------- m/s                                                   = ---------- m/s             

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