A weightless spring is stretched 10 cm by a suspended 1-kg block.If two such springs are used to suspend the block,one spring above the other,to effectively provide one double-length spring,then the total stretch of the double-length spring will be
A)5 cm.
B)10 cm.
C)20 cm.
D)30 cm.
E)none of these

Question

Quizplus · Accepted Answer

The spring constant of a spring is given by k = F/x, where F is the force applied and x is the displacement or stretch. Therefore, if one spring is stretched by 10 cm due to the weight of a 1-kg block, the spring constant of that spring can be found as follows:

k1 = F/x = (1 kg) * (9.8 m/s^2) / (0.1 m) = 98 N/m.

When two identical springs are used in parallel, the effective spring constant is simply the sum of the individual spring constants, i.e., k_eff = 2k1 = 196 N/m. The stretch of the double-length spring can be found by using Hooke's law:

F = k_eff * x_eff,

where x_eff is the stretch of the double-length spring. When the 1-kg block is suspended using the double-length spring, the force applied is simply the weight of the block, i.e., F = (1 kg) * (9.8 m/s^2) = 9.8 N. Therefore,

x_eff = F / k_eff = (9.8 N) / (196 N/m) = 0.05 m = 5 cm.

Therefore, the total stretch of the double-length spring is 2*x_eff = 2*(5 cm) = 10 cm.

A Weightless Spring Is Stretched 10 Cm by a Suspended