(c7p50) A 275-g block is dropped onto a vertical spring with spring constant k =

Question

(c7p50) A 275-g block is dropped onto a vertical spring with spring constant k =110.0 N/m. The block becomes attached to the spring, and the spring compresses 38.2 cm before momentarily stopping. While the spring is being compressed, a)what work is done on the block by its weight?b)While the spring is being compressed, what work is done on the block by 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

Spring constant = k = 110.0 N/cm = 110000 N/m

Mass = m=275 g =0.275 kg

The compression of spring = x =38.2 cm

Work done on the block by the gravitational force =mgh =0.275*9.8*38.2

a) work done on the block by gravitational force = 102.949 J

b) Work done on the block by the spring force while the spring is being compressed=(1/2)kx^2 = 0.5*110000*0.382*0.382 = 8025.82 J

speed of the block just before it hits the spring=v =[ sq rt2*KE/m]

v = sq rt [2*( 102.949 +8025.82)/0.275]

v =243.14 m/s

c)The speed of the block just before it hits the spring is 243.14 m/s

If speed is doubled ,the maximum compression is twice the initial compression because,

(1/2)mv^2=(1/2)kx^2

x is proportional to v

d) If the speed at impact is doubled, the maximum compression of the spring will be 76.4 cm

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