A 250 g block is dropped onto a relaxed vertical spring that has a spring consta

Question

A 250 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2.7 N/cm. The block becomes attached to the spring and compresses the spring 14 cm before momentarily stopping. While the spring is being compressed, what work is done on the block by the gravitational force on it? J What work is done on the block by the spring force while the spring is being compressed? J What is the speed of the block just before it hits the spring? (assume that friction is negligible. ) m/s If the speed at impact is doubled, what is the maximum compression of the spring? m

Explanation / Answer

change the mass to 250g and stiffness to 2.7N/cm initial displacement =14cm Spring constant = k = 3.0 N/cm = 300 N/m Mass = m=270 g =0.27 kg The compression of spring = x = 11 cm Work done on the block by the gravitational force =mgh =0.27*9.8*0.11 a) work done on the block by gravitational force =0.29106 J ______________________________________… b) Work done on the block by the spring force while the spring is being compressed=(1/2)kx^2 = 0.5*300*011*0.11 b) Work done on the block by the spring force while the spring is being compressed=1.815 J speed of the block just before it hits the spring=v =[ sq rt2*KE/m] v = sq rt [2*(0.29106+1.815)/0.27] v = sq rt [15.6004] v =3.9497 m/s c)The speed of the block just before it hits the spring is 3.9497 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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