A 2.00kg block is attached to a spring of force constant 500N/m. The block is pu

Question

A 2.00kg block is attached to a spring of force constant 500N/m. The block is pulled 5.0 cm to the right of equilibriumand released. Find the speed of the block as it passes throughequilibrium if a) The horizontal surface is frictionless? b) The coefficient of friction between the block and surfaceis 0.35 A 2.00kg block is attached to a spring of force constant 500N/m. The block is pulled 5.0 cm to the right of equilibriumand released. Find the speed of the block as it passes throughequilibrium if a) The horizontal surface is frictionless? b) The coefficient of friction between the block and surfaceis 0.35

Explanation / Answer

a) If there's no friction just set the potential energy equalto the kinetic energy. (1/2)mv2=(1/2)kx2 (1/2)(2)v2=(1/2)(500)(.05)2
v=1.12 m/s
b) Do the same thing but add frictional energy to the finalenergy (1/2)kx2=(1/2)mv2+mgx (1/2)(500)(.052)=(1/2)(2)v2+(2)(9.8)(.35)(.05) v=0.531 m/s b) Do the same thing but add frictional energy to the finalenergy (1/2)kx2=(1/2)mv2+mgx (1/2)(500)(.052)=(1/2)(2)v2+(2)(9.8)(.35)(.05)

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