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

#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'}#

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'#