A 100- g block is dropped onto a vertical spring with spring constant k=2.0 N/cm

Question

A 100-g block is dropped onto a vertical spring with spring constant k=2.0 N/cm. The block becomes attached to the spring, and the spring compresses 12.0 cm before momentarily stopping.

While the spring is being compressed, what work is done by the block's weight?

While the spring is being compressed, what work is done by the spring?

What was the speed of the block just before it hits the spring?

If the speed of impact is doubled, what is the maximum compression of the spring? Assume the friction is negligible

Explanation / Answer

Assume that the block is dropped from a height h onto the spring.

work done by block's weight = mg (h + 0.12)

work done by spring = - (1/2) x (200) x (0.12)2 = - 1.44 J

speed of block just before it hits the spring = (2 x 9.8 x h)1/2

using work-energy theorem :: work done by all forces = change in kinetic energy

mg (h + 0.12)  - (1/2) x (200) x (0.12)2 = 0

h = 1.35 m

to double the speed of impact, we have to release it from a height H

2 x (2gh)1/2 = (2gH)1/2

H = 4h = 5.4 m

let the maximum compresion be S

mg (H + S)  - (1/2) x (200) x (S)2 = 0

mg (4h + S)  - (1/2) x (200) x (S)2 = 0

mg 4h = (1/2) x (200) x (S)2 - mgS

since mg h = (1/2) x (200) x (0.12)2 - mg x 0.12

Hence 4 x ( (1/2) x (200) x (0.12)2 - 0.98 x 0.12 ) = 100 S2 - 0.98S

100 S2 - 0.98 S - 5.2896 = 0

S = 0.235 m

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