A 250 g block is dropped onto a relaxed vertical spring that has a spring consta

Question

A 250 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2.5 N/cm is shown in figure below. The block becomes attached to the spring and compresses the spring 12 cm 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.) (d) If the speed at impact is doubled, what is the maximum compression of the spring?

Explanation / Answer

We have:

Spring constant = k = 2.5 N/cm = 0.025 N/m

Mass = m = 250 g = 0.25 kg

The compression of spring = x = 12 cm

(A) Work done on the block by the gravitational force = mgh = 0.25x 9.8 x 0.12 = 0.294 J

So, The work done on the block by gravitational force is 0.294 J.

(B) Work done on the block by the spring force while the spring is being compressed is

= (1/2) kx2 = 0.5 x 0.025 x (0.12)2

So, Work done on the block by the spring force is 1.8 x 10-4 J.

(C) The speed of the block just before it hits the spring=v =[ sqrt 2*KE/m]

v = sqrt [2*(0.294 +1.8 x 10-4) / 0.25]

v = sqrt [4.1692]

v = 2.169 m/s

So, The speed of the block just before it hits the spring is 2.169 m/s.

(D) If speed is doubled ,the maximum compression is twice the initial compression because,

(1/2)mv2=(1/2)kx2

x is proportional to v

So, If the speed at impact is doubled, the maximum compression of the spring will be 24 cm or 0.24 m.

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