A 1.70-kg object is hanging from the end of a vertical spring. The spring consta

Question

A 1.70-kg object is hanging from the end of a vertical spring. The spring constant is 54.0 N/m. The object is pulled 0.200 m downward and released from rest. Complete the following table by calculating the translational kinetic energy, the gravitational potential energy, the elastic potential energy, and the total mechanical energy E for each of the vertical positions listed. The vertical positions h indicate distances above the point of release, where h = 0 m.

Explanation / Answer

At h=0 The spring is in stretched position x0 such that mg=kx0

solving this we get x0=0.308m

At maximum stretch total elongation in spring is 0.508m.(xf)

Elastic Potential energy =(1/2)X k X xf2 = 6.9677 joule.

At xf K.E.= 0; P.E.= 0( Datum)

At xo P.E.(gravitation)= mXgX0.2 = 3.3354 joule

P.E.(spring) =(1/2)X k X xo2=2.5613 joule.

So, K.E.= P.E.(Spring)-P.E.(gravitation)=1.071

Total energy = [P.E.(spring)+P.E.(gravitation)+K.E.]0=  [P.E.(spring)+P.E.(gravitation)+K.E.]f= 6.9677 j.

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