A 320 g block is dropped onto a relaxed vertical spring that has a spring consta

Question

A 320 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2.7 N/cm. The block becomes attached to the spring and compresses the spring 19 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?

For a) and b) I would suppose they're equal and opposite (Newton's 3rd), and that I can just take the Work W = 2.7N/.02M(.019M^2) = .48735J, but that's not correct apparently.

For c) I tried to use kinematics, Vi^2 = Vf^2 - 2ad => Vi = sqrt(0-s*9.81m/s^2*.019m)
=> 0.61055 m/s, but this is also wrong. Some help? Perhaps?

Explanation / Answer

The mass of the block m = 0.32kg the spring constant k = 2.7 N/cm = 270 N/m the compression of the spring x = 0.19m (a) The gravitational force            F = mg = (0.32) (9.8)                        = 3.13 N (b) Spring force F = - 3.13 N According to the Newton's third law (c) From law of conservation of enregy        mg(x0-x) + 1/2 kx^2 = 1/2mv^2 + mgx0     mgx0 - mgx + 1/2 kx^2 = 1/2mv^2 + mgx0      - mgx + (0.5) kx^2 = (0.5) mv^2     - (0.32) (9.8)(0.19) + (0.5)(270)(0.19)^2 = (0.5)(0.32)v^2 -0.595 + 4.874 = 0.16 v^2 therefore the speed                 v = 5.2 m/s (d) If the speed of impact is doubled then   - mgx + (0.5) kx^2 = (0.5) mv^2            - (0.32) (9.8) x + (0.5)(270) x^2 = (0.5)(0.32)(10.4)^2             135 x^2 - 3.14x - 17.3 = 0 therefore the compression         x = 0.369 or 37cm      - mgx + (0.5) kx^2 = (0.5) mv^2     - (0.32) (9.8)(0.19) + (0.5)(270)(0.19)^2 = (0.5)(0.32)v^2 -0.595 + 4.874 = 0.16 v^2 therefore the speed                 v = 5.2 m/s (d) If the speed of impact is doubled then   - mgx + (0.5) kx^2 = (0.5) mv^2            - (0.32) (9.8) x + (0.5)(270) x^2 = (0.5)(0.32)(10.4)^2             135 x^2 - 3.14x - 17.3 = 0 therefore the compression         x = 0.369 or 37cm

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