a 2 kg block is attached to a spring of force constant 440N/m. the block is pull
ID: 1740918 • Letter: A
Question
a 2 kg block is attached to a spring of force constant 440N/m. the block is pulled 5.20 cm to the right of equilibrium andreleased from rest. a. find the speed of the block as it passes throughequilibrium if the horizontal surface is frictionless b. find the speed of the block as it passes throughequilibrium (for the first time) if the coefficient of frictionbetween the block and the surface is 0.350 a 2 kg block is attached to a spring of force constant 440N/m. the block is pulled 5.20 cm to the right of equilibrium andreleased from rest. a. find the speed of the block as it passes throughequilibrium if the horizontal surface is frictionless b. find the speed of the block as it passes throughequilibrium (for the first time) if the coefficient of frictionbetween the block and the surface is 0.350Explanation / Answer
we are given with m = 2.0 kg k = 440 N / m xi = 0.0520 cm (a) when the friction is not present the energyconservation gives (KE + PEg +PEs)f = (KE + PEg +PEs)i (1 / 2) m vf2 + 0 + 0 = 0 +0 + (1 / 2) k xi2 so the speed of the block willbe vf = xi (k /m) = ........ cm /s (b) when the friction is present thework energy thm gives Wnc = (KE + PEg +PEs)f - (KE + PEg +PEs)i - f . xi = [(1 / 2) mvf2 + 0 + 0] - [0 + 0 + (1 / 2) kxi2] vf = ....... cm / s (c) when the final velocity ios zero at x = 0then - f . xi = (0) - [0 + 0+ (1 / 2) k xi2] then the strength of the frictional force willbe f = k xi / 2 = ........ N so the speed of the block willbe vf = xi (k /m) = ........ cm /s (b) when the friction is present thework energy thm gives Wnc = (KE + PEg +PEs)f - (KE + PEg +PEs)i - f . xi = [(1 / 2) mvf2 + 0 + 0] - [0 + 0 + (1 / 2) kxi2] vf = ....... cm / s (c) when the final velocity ios zero at x = 0then - f . xi = (0) - [0 + 0+ (1 / 2) k xi2] then the strength of the frictional force willbe f = k xi / 2 = ........ N
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