Question

Two springs, with force constants k1=175N/m and k2=270N/m, are connected in series

When a mass m=0.50kg is attached to the springs, what is the amount of stretch, x?
F = M*g = 0.50 * 9.8 = 4.9 N.

X = (1m/175N)*4.9N. + (1m/270N)*4.9N = 0.028 + 0.0181 = 0.0461 m.


Related Questions

The figure below shows two springs connected in parallel. This combination can be thought of as bein... Two springs, with force constants k1=175N/m and k2=270N/m, are connected in series, as shown in (Fig... Two springs, with force constants k1=175N/m and k2=270N/m, are connected in series, as shown in (Fig... When a force of 36 Newtons is applied to springs S1 and S2, the displacement of the springs is 6 cen...