A 217 kg block is released at a 6.6 m height as shown. The track is frictionless

Question

A 217 kg block is released at a 6.6 m height as shown. The track is frictionless except for a portion of length 7.3 m . The block travels down the track, hits a spring of force constant k = 1916 N/m . The coefficient of kinetic friction between surface and block over the 7.3 m track length is 0.56 . The acceleration of gravity is 9.8 m/s2 . Determine the compression of the spring x from its equilibrium position before coming to rest momentarily.Answer in units of m.

Note: Please show work and good explanation

Explanation / Answer

We solve this by using conservation of energy...

Step 1: As the path is frictionless till the green patch, no work is lost to friction.

Hence the P.E at top rest point is equal to the K.E just before the patch which is

P,E=K.E= m.g.h= 217x9.8x6.6 = 14035.56 Joule

Step 2:

In the friction path, work done by friction force 'f' is(S is the distance travelled in friction)

f= 0.56 x m.g

Wfric = f.S = 0.56 x 217x9.8 x 7.3 = 8693.54 Joule

Step 3: The lost work in friction is deducted and the rest of the work goes into spring energy.

Wremaining = 5342.02 Joule = Spring energy = 0.5xk.x^2 = 0.5x1916 x x^2

x^2 = 5.5762 m

x = 2.361 m

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