A) A 200-g block is dropped onto a vertical spring with spring constant k =220.0

Question

A) A 200-g block is dropped onto a vertical spring with spring constant k =220.0 N/m. The block becomes attached to the spring, and the spring compresses 29.5 cm before momentarily stopping. While the spring is being compressed, 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 = 3.0 N/cm = 300 N/m Mass = m=200 g =0.2 kg The compression of spring = x = 29.5 cm Work done on the block by the gravitational force =mgh =0.2*9.8*0.295 a) work done on the block by gravitational force =0.5782J ______________________________________… b) Work done on the block by the spring force while the spring is being compressed=(1/2)kx^2 = 0.5*220*0.295^2 b) Work done on the block by the spring force while the spring is being compressed=9.57J speed of the block just before it hits the spring=v =[ sq rt2*KE/m] v = sq rt [2*(0.5782+9.57)/0.2] v = sq rt [15.6004] v =10.07 m/s c)The speed of the block just before it hits the spring is 10.07 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 22 cm

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