A mass of 4.5 kg,resting on a horizontal frictionless surface,is attached on the right to a horizontal spring with spring constant 29 N/m and on the left to a horizontal spring with spring constant 60 N/m.If this system is moved from equilibrium,what is the effective spring constant?
A)-31 N/m
B)31 N/m
C)89 N/m
D)-89 N/m
E)16 N/m

Question

Quizplus · Accepted Answer

The effective spring constant of two springs in parallel is given by:
1/k_eff = 1/k1 + 1/k2
where k1 and k2 are the spring constants of the individual springs. 
Using this formula, we get:
1/k_eff = 1/29 + 1/60
1/k_eff = (60 + 29)/(29*60)
k_eff = 89 N/m (since k_eff > 0, the negative sign in option A and D is incorrect)

A Mass of 4