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Question

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2. (20) A spring in figure 2a rest on a 30 degree incline, Its uncompressed length is 0.8m. A 2.0 kg mass is placed on the incline and pushed down so that the spring is compressed 0.6 m from it's equilibrium position. If the coefficient of kinetic friction between the mass and the incline is 0.2 and the block moves up the incline 1.0 m after the spring is released determine the spring constant k.

Explanation / Answer

The weight of the mass, W = 2*(9.8) =19.6 N. Lets divide it along two components, Perpendicular to the incline, W*cos 30, which is also equal to the normal reaction force applied by the plane.
                        R= W*cos30                           = 19.6cos30                           =16.97N
Frictional force, fr = R* 0.2                              = 3.39 = 3.4N
Component of weight along the plane                W*sin 30 =9.8N
Net force along the plane,                      Fnet = 9.8 + 3.4 =13.2 N
therefore work done to move the spring                   W = F*s = 13.2*(1m)                                = 13.2 J
which was given as kinetic energy by the spring and was stored as spring potential energy in the spring.
therefore, .5*k*x^2= 13.2 J
or,             k= 13.2/(.5* 0.6^2)                    =73.33 N/m

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