A spring stretches 6.0 cm when a 0.25 kg block is hung from it.?

A spring stretches 6.0 cm when a 0.25 kg block is hung from it.

If a 0.60 kg block replaces the 0.25kg block, how far does the spring stretch?

• By Hooke's law, the force, F, to stretch a spring a distance of d is:

F = kd, where k is a constant.

Since 0.25*9.8 = 2.45 N of force stretches the spring a distance of 6.0 cm:

2.45 = 6.0k ==> k = 2.45/6.

Then, when F = 0.60*9.8 = 5.88 N:

5.88 = (2.45/6)d ==> d = [(5.88)(6)]/2.45 = 14 cm, to 2 s.f.

I hope this helps!

• 6.0 In Cm

Source(s): https://owly.im/a95C4
• The pressure on both.a million kg (= m1) block is m1*g + ok*x1 = 0 before the 315 g (=m2) mass is further. After m2 is further the pressure on block m1 is (m1+m2)g + ok(x1+x2) = 0. So we've a gadget of two equation and 2 unknowns, ok and x1: m1*g + ok*x1 = 0 (m1+m2)g + ok(x1+x2) = 0 ascertain for ok and compute. For a spring mass gadget the angular frequency w=(ok/m)^(a million/2)=2*pi/T, ascertain for T and compute.