A block of mass m is dropped onto a relaxed vertical spring with spring constant

Question

A block of mass m is dropped onto a relaxed vertical spring with spring constant k (see the figure). The block becomes attached to the spring and compresses the spring a distance d before momentarily stopping. While the spring is being compressed, what work is done on the block by (a) the gravitational force on it and (b)the spring force? (c) What is the speed of the block just before it hits the spring? (Assume that friction is negligible.) State your answers in terms of the given variables and g.

Explanation / Answer

(a) The work by the gravitational force on the block is

               Wgrav = mgd cos(0) = mgd

(b) The work done by the spring force is,

                      Wspring   (= 1/2 kx _{i}^{2} - 1/2 kx _{f}^{2})               

(c) The workenergy theorem says that the change in the kinetic energy is equal to

                the total work done:

                   (Delta) KE = Wgrav +Wspring

                                         (Delta K = 1/2 m v_{f}^{2} - 1/2 m v_{i}^{2})

If the spring has initial velocity vi and is at rest finally , then vf =0, then,

                         (0 - 1/2 m v_{i}^{2} = W_{grav} + W _{spring})    

                                                  (v_{i}^{2} = 2 ( W_{grav} + W_{spring} ) / m)     

                                                  (v_{i} = sqrt{2(W_{grav} + W_{spring}) / 2})

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