# 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?

${F}_{s} = \setminus \frac{k k ' \left(x + x '\right)}{k + k '}$

#### Explanation:

The equivalent stiffness (${k}_{e}$) of spring when two springs of stiffness $k$ & $k '$ are connected in series is given as

$\frac{1}{k} _ e = \frac{1}{k} + \frac{1}{k '}$

${k}_{e} = \setminus \frac{k k '}{k + k '}$

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

${F}_{s} = {k}_{e} {x}_{e}$

${F}_{s} = \setminus \frac{k k '}{k + k '} \left(x + x '\right)$

${F}_{s} = \setminus \frac{k k ' \left(x + x '\right)}{k + k '}$

Jun 27, 2018

$F = k x = k ' x '$

#### Explanation:

The most importat thing to understand here is that the force $\left(F\right)$ 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":

• $F = k x$

• $F = k ' x '$