A 240 g block is dropped onto a relaxed vertical spring that has a spring consta
ID: 2258198 • Letter: A
Question
A 240 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2.6 N/cm (see the figure). The block becomes attached to the spring and compresses the spring 13 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) 0.30576 J
Explanation / Answer
Spring constant = k = 2.6 N/cm = 260 N/m
Mass (m) = 240 g = 0.24 kg
The compression of spring = x = 13 cm = 0.13 m
(a)
Work done on the block by the gravitational force = mgh = 0.24*9.8*0.13 = 0.30576 J
b)
Work done on the block by the spring force while the spring is being compressed
= (1/2)kx^2
= 0.5*260*(0.13)^2
= 2.197 J
(c)
speed of the block just before it hits the spring = v =[ sqrt(2*KE/m)]
v = sqrt [2*(0.30576 + 2.197)/0.24]
v = sqrt [20.8563]
v = 4.57 m/s
(d)
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
the maximum compression of the spring = 13*2 = 26 cm
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