Question Solved1 Answer A 2.00-kg block is attached to a spring of force constant 500 N/m. The block is pulled 5.00 cm to the right of equilibrium and released from rest. Find the speed of the block as it passes through equilibrium if (a) the horizontal surface is frictionless and (b) the coefficient of friction between block and surface is 0.350

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Transcribed Image Text: A 2.00-kg block is attached to a spring of force constant 500 N/m. The block is pulled 5.00 cm to the right of equilibrium and released from rest. Find the speed of the block as it passes through equilibrium if (a) the horizontal surface is frictionless and (b) the coefficient of friction between block and surface is 0.350
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Transcribed Image Text: A 2.00-kg block is attached to a spring of force constant 500 N/m. The block is pulled 5.00 cm to the right of equilibrium and released from rest. Find the speed of the block as it passes through equilibrium if (a) the horizontal surface is frictionless and (b) the coefficient of friction between block and surface is 0.350
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Solution rarr(a) rarr From the conservation of energy rarr{:[(1)/(2)mV^(2)=(1)/(2)KA^(2)],[=>2.00V^(2)=500 xx(0.05)^(2)],[[As" Amplitude "=A=0.05m]]:}So V^(2)=(1.25)/(2.00)=0.625{:[OR=0.7906m//s],[V=0.7006m//s]:}(b) f_(k)=mu_(k)Mg=0.350 xx2.00 xx9.8 ... See the full answer