A relaxed spring of length 0.19 m stands vertically on the floor; its stiffness is 1270 N/m. You release a block of mass 0.5 kg from rest, with the bottom of the block 0.7 m above the floor and straight above the spring. How long is the spring when the block comes momentarily to rest on the compressed spring? m
Gravitational acceleration = g = 9.81 m/s2
Mass of the block = m = 0.5 kg
Force constant of the spring = k = 1270 N/m
Length of the spring = L0 = 0.19 m
Height of the block above the floor = H0 = 0.7 m
Initial height of the block from the top of the spring = H
H = H0 - L0
H = 0.7 - 0.19
H = 0.51 m
Compression of the spring when the block comes momentarily to rest = X
The initial potential energy of the block is converted into the potential energy of the spring as the block comes to rest on the spring.
mg(H + X) = kX2/2
(0.5)(9.81)(0.51 + X) = (1270)X2/2
2.502 + 4.905X = 635X2
635X2 - 4.905X - 2.502 = 0
X = 0.0667 m or -0.059 m
Distance cannot be negative.
X = 0.0667 m
Length of the spring when the block comes momentarily to rest = L
L = L0 - X
L = 0.19 - 0.0667
L = 0.1233 m
Length of the spring when the block comes momentarily to rest on the compressed spring = 0.1233 m
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