Question

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?

Answer 1

`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'}`

Answer 2

`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'`