A 230 g block is dropped onto a relaxed vertical spring that has a spring consta

Question

A 230 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 3.9 N/cm (see the figure). The block becomes attached to the spring and compresses the spring 17 cm before momentarily stopping. While the spring is being compressed, what work is done on the block by (a) the gravitational force on it and (b) 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 = 3.9 N/cm = 390 N/m

Mass = m=230 g =0.23 kg

The compression of spring = x = 17 cm

Work done on the block by the gravitational force =mgh =0.23*9.8*0.17=0.3836J

a) work done on the block by gravitational force =0.3836 J
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b) Work done on the block by the spring force while the spring is being compressed=(1/2)kx^2 = 0.5*390*0.17^2=5.6355J

b) Work done on the block by the spring force while the spring is being compressed=5.6355J

speed of the block just before it hits the spring=v =sqrt(2*KE/m)

v = sqrt [2*(0.3836+5.6355)/0.23]

v^2 = 52.34
v =7.235 m/s

c)The speed of the block just before it hits the spring is 7.235m/s
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If speed is doubled ,the maximum compression is twice the initial compression because,

(1/2)mv^2=(1/2)kx^2+m*g*x

=>0.5*0.23*14.47^2 =0.5*390*x^2+2.2563*x

=>195*x^2-2.2563*x-24.079=0

=>x=0.36m ..........(c)

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