A 2.45 kg puck is placed against a spring of spring constant 150 N/m and compres

Question

A 2.45 kg puck is placed against a spring of spring constant 150 N/m and compressed 0.16 m (point A). It is released (leaves spring at point B) and slides across a frictionless surface until it encounters (point C) a portion of the surface that is 0.26 m long with a coefficient of kinetic friction of 0.20. After leaving that surface (at point D), it slides until it strikes (at point E) a spring attached to a support with a spring constant of 100 N/m. The spring compresses to point F.

How far is the second spring compressed?

Explanation / Answer

The mass of the puck m = 2.45kg

the spring constant k1 = 150N/m

the compression x1 = 0.16m

the spring constant k2 = 100N/m

the coefficient of friction = 0.2

From law of conservation of energy

    1/2k1x1^2 = mg*d + 1/2k2x2^2

    k1x1^2 = 2 mg*d + k2x2^2

    k2x2^2  =  k1x1^2 - 2 mg*d 

therefore the compression of the spring

     x2 = (k1x1^2 - 2mg*d)/k2

         = [(150)(0.16)^2 - 2(0.2)(2.45)(9.8)(0.26)]/100

          =  0.013 m or 1.3cm

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