Two blocks of mass 4kg and 6kg are attached by a spring constant k=200N/m, both blocks are moving with same acceleration. Find elongation in spring
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Hence, extension of spring is 30cm
Explanation:
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Given:
Mass of two blocks = 4 kg and 6
Spring constant k = 200 N/m
To find:
Elongation in the spring if the two blocks are moving with same acceleration.
Solution:
Let the elongation in the spring = 6 kg
And the acceleration be a m/s^2
For 6 kg block, as observed from the FBD the equation will be:
20 - kx = 6a
(1)
For the 4 kg block, the equation will be:
kx = 4a (2)
On adding both the equations we get:
20 = 10a
a = 2 m/s^2
From equation 2:
kx = 4a
x=0.04m
Hence, elongation of the spring is 2 m.
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