A 1.60 kg block slides with a speed of 0.905 m/s on a frictionless horizontal su

Question

A 1.60 kg block slides with a speed of 0.905 m/s on a frictionless horizontal surface until it encounters a spring with a force constant of 761 N/m. The block comes to rest after compressing the spring 4.15 cm. Find the spring potential energy, PE, the kinetic energy of the block, KE, and the total mechanical energy of the system, E, for compressions of 1cm .

From my math, I'm getting KE= 1/2mv^2= (1/2)(1.60kg)(.905m/s)^2= .655 J
PE= 1/2kx^2= (1/2)(761N/m)(.01m)^2= .038 J and
E= PE + KE, at 0 compression PE=0, so E= .038J + .655J= .693 J
But, obviously, this answer is incorrect. Any ideas?

Explanation / Answer

Kinetic energy before compression(zero compression) and after compressing to 1 cm will be different.So in calculating the total energy at compression 1 cm you cannot take it granted that kinetic energy is 0.655J as the velocity changes. However as there are no non conservative forces.So the total energy is conserved and remains constant at any time. Total Energy at compression 1 cm = Total energy before compression = K.E.+P.E. T =0.655 + 0 =0.655 J But Spring potential energy at compression 1 cm = (1/2)*k*(0.01)^2 = 0.038 J At this compression total energy T = (KE + PE) at this compression = KE + 0.038 But T = 0.655 KE = 0.655-0.038 = 0.617 J Hence the kinetic energy = 0.617 J

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