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