A 260 g block is dropped onto a relaxed vertical spring that has a spring consta

Question

A 260 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 3.0 N/cm (Fig. 7-30). The block becomes attached to the spring and compresses the spring 11 cm before momentarily stopping.

(a) While the spring is being compressed, what work is done on the block by the gravitational force on it?
J
(b) What work is done on the block by the spring force while the spring is being compressed?
J
(c) What is the speed of the block just before it hits the spring? (Assume that friction is negligible.)
m/s
(d) If the speed at impact is doubled, what is the maximum compression of the spring?
m

Thank You!

Explanation / Answer

m = 260 g = 0.26 kg

k = 3 N/cm = 300 N/m

total compression = 11 cm = 0.11 m

So energy stored in spring at maximum compression = 0.5kx2 = 0.5*300*0.112 = 1.815 J

energy gained due to lowering of 11 cm height = mhg = 0.26 * 0.11 * 9.81 = 0.280566 J

a) work is done on the block by the gravitational force on it? = mgh = 0.280566 J

b) work is done on the block by the spring force while the spring is being compressed = - 0.5kx2 = - 1.815 J

c) Let the speed of block just before it hits spring be v

Then by conservation of energy

0.5mv2 + mgh = 0.5kx2

=> 0.5*0.26*v2 + 0.280566 = 1.815

=> v = 3.4356 m/s

the speed of the block just before it hits the spring = 3.4356 m/s

d) new velocity = 2 * 3.4356 = 6.871 m/s

0.5mv2 + mgh = 0.5kx2

=> 0.5*0.26*6.8712 + 0.26*9.81*0.11 = 0.5 * 300 * x2

=> 150 x2 = 6.4179

=> x = 0.2068 m = 20.68 cm

So new maximum compression = 20.68 cm

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