Two springs are in a series combination and are attached to a block of mass 'm' which is in equilibrium. Spring constants:k and k’ Extensions:x and x’ respectively. Find the force exerted by spring on the block?

2 Answers
Harish Chandra Rajpoot
Jun 26, 2018

F_s=\frac{kk'(x+x')}{k+k'}

Explanation:

The equivalent stiffness (k_e) of spring when two springs of stiffness k & k' are connected in series is given as

1/k_e=1/k+1/{k'}

k_e=\frac{kk'}{k+k'}

Now, the total extension in the equivalent spring
x_e=x+x'

hence, the force (F_s) in each of the springs i.e. force in the equivalent spring

F_s=k_ex_e

F_s=\frac{kk'}{k+k'}(x+x')

F_s=\frac{kk'(x+x')}{k+k'}

Ultrilliam
Jun 27, 2018

F = kx = k'x'

Explanation:

The most importat thing to understand here is that the force (F) is the same in each spring, in the same way that the tension in a chain is the same in each link of that chain.

This then leads to 2 very simple expressions for "the force exerted by spring on the block":

  • F = kx

  • F = k'x'

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