A 525-g block is dropped onto a vertical spring with spring constant k =200.0 N/

Question

A 525-g block is dropped onto a vertical spring with spring constant k =200.0 N/m. The block becomes attached to the spring, and the spring compresses 54.9 cm before momentarily stopping. While the spring is being compressed, what work is done on the block by its weight?

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

What is the speed of the block just before it hits the spring? (Assume that friction is negligible.)

If the speed at impact is doubled, what is the maximum compression of the spring?

Explanation / Answer

(a) Work done by weight on block = weight x displacement = mgx = 0.525 kg x (-9.81 m/s^2)x (-0.549 m) = 2.8 J

(b) Work done by spring force = -( change in spring potential energy) = -( final spring potential energy - initial spring potential energy) = - ( 0.5kx^2 - 0) = - (0.5 x 200 x 0.549^2) = -30.14 J

(c) Let speed of block just befor hitting the spring = v

Initial kinetic energy of block + Initial potential energy of block = Final spring potential energy

0.5 mv^2 + mgx = 0.5kx^2

0.5( 0.525)v^2 + (0.525)(9.81)(0.549) = 0.5(200)(0.549)^2

v = 10.2 m/s

(d) If v is doubled to 20.4 m/s, then

0.5 mv^2 + mgx = 0.5kx^2

0.5( 0.525)(20.4)^2 + (0.525)(9.81)x= 0.5(200)(x)^2

109.24 + 5.15x = 100x^2

100x^2 -5.15x - 109.24 = 0

x = 1.07 m

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