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Physics 22 Online
OpenStudy (anonymous):

A 3.5 kg block is accelerated from rest by a compressed spring of spring constant 640 N/m. The block leaves the spring at the spring's equilibrium length and then immediately begins to travel over a horizontal floor with coefficient of kinetic friction μk = 0.25. The block stops in a distance D = 7.8 m. What was the original compression distance of the spring? Not really sure how to approach this one. Thanks guys

OpenStudy (anonymous):

when the spring is compressed by "x", it stores potential energy of \(U_s={1\over2}kx^2\) This potential energy gets transferred to kinetic energy of the object. Work is done by friction that makes final energy=0 i.e., W=entire initial kinetic energy.

OpenStudy (anonymous):

electrokoid is right, as 1/2kx^2=work where work=f*d=force of friction times distance, that it 1/2kx^2=mk*N*d sustituting constants 1/2*640*x^2=0.25*3.5*9.8 (gravity)* 7.8 x square = 4.78, x equals 2.187meters.

OpenStudy (anonymous):

sorry i've missed sth, at conservation of energy, but at whole equation will look like this

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