A block of mass M is dropped onto the top of a vertical spring whose force const

Question

A block of mass M is dropped onto the top of a vertical spring whose force constant is k. The block is released from a height H above the top of the (relaxed) spring. Define u to be the displacement of the spring below its equilibrium point (i.e. when the block is in contact with it).

3.2 In this question we will graph the kinetic energy vs. displacement. One of the ways used to draw general conclusions from specific situations is to use appropriate combinations of variables. First, write the expression for the kinetic energy, K, in terms of M, g, H, k and u. Next, rewrite the expression as a ratio: kappa = K/(MgH). Next substitute in your expression for kappa the (again dimensionless) variable w = u/H. Graph the 'reduced' kinetic energy, kappa, as a function of w. There will be one other dimensionless combination appearing in your result: kH/Mg which relates the spring force to the weight. Make your graph for kH/Mg = 3. Verify that the maximum value of kappa appears in the correct place.

Explanation / Answer

Since you have all the values so graph can be drawn. So i am describing here how to calculate displacement u.

The total height through which the block falls = H + u            ( u = displacement by the spring compressed)

The loss in potential energy of the body = Mg ( H+ u)

The energy stored in the spring = 1/2* k u2

Since the potential energy of block is used to compress the spring .

Mg ( H+u) = 1/2*k*u2

By putting the values of mass of block = M, height H through which it has fallen , spring constant K we can calculate displacement of spring u.

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