An object of mass 2.35 kg is traveling on a horizontal surface. There is a coeff

Question

An object of mass 2.35 kg is traveling on a horizontal surface. There is a coefficient of kinetic friction 0.206 between the object and the surface. The object has speed 2.49 m/s when it reaches x=0 and encounters a spring. The object compresses the spring, stops, and then recoils and travels in the opposite direction. When the object reaches x=0 on its return trip, it stops.
1) Find k, the spring constant
These are the hints that are given:a) Why does the object stop? b) energy stored in a spring c) compute the compression of the spring d) putting it all together e) the value of E final f) the answer is in N/m

Explanation / Answer

According to conservation of energy, Initial : 1/2mv2 = Wf +Es here Wf = f. x     (work done against friction ) here Wf = f. x     (work done against friction )              = mg.x and Es = 1/2k.x2 (Energy stored in the spring)      1/2mv2 = mg.x +  0.5k.x2 -----------(1) On return, final:           1/2 k.x2 =  mg.x  -----------------(2) From (1) and (2)               1/2mv2 = 2 mg.x                 k = 2mg./x Solving between theabove 2 eqns,               k * 1/2mv2 = (2mg.)2                  k =  (2mg.)2/0.5mv2                     = 8 m(g/v)2                          = 8*2.35(0.206*9.8/2.49)2                     =12.35 N/m                     =12.35 N/m

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