Chapter 07, Problem 062 A 310 g block is dropped onto a relaxed vertical spring

Question

Chapter 07, Problem 062 A 310 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2.2 N/cm (see the figure). The block becomes attached to the spring and compresses the spring 16 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? (a) Number Units (b) Number Units (c) Number Units (d) Number Units

Explanation / Answer

A)

The compression of the spring is d = 0.12 m. The work done by the force of gravity (acting on the
block) is, by Eq
W1 = mgd = (0.310 kg) (9.8m/s2) (0.16 m) = 0.486 J

B)

The work done by the spring is, by Eq

W2 = ?1/2 kd2 = ?1/2(220N/m)(0.16 m)2 = ?2.816 J

C)

The speed vi of the block just before it hits the spring is found from the work-kinetic energy theorem
(Eq.

?K = 0 ? 1/2 mv2i = W1 +W2

which yields

vi = sqrt [(?2)(W1 +W2)/m] = sqrt [(?2)(0.486 ? 2.816)/0.31] = 3.87 m/s

D)

If we instead had vi = 7.74 m/s, we reverse the above steps and solve for d. Recalling the theorem
used in part (c), we have

0 ? 1/2 mv'i2 = W'1 + W'2

= mgd' ? 1/2 kd' 2

which (choosing the positive root) leads to

d' = mg + sqrt [m2g2 + mkvi2]/k

= 0.310 + sqrt [ (0.310)2(9.8)2 + 0.310)2 (220) (3.87)2]/220 = 1.52

which yields d = 1.52 m. In order to obtain this, we have used more digits in our intermediate
results than are shown above (so vi = ?15.03 = 3.87 m/s and vi' = 7.74 m/s).

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