Question

7. State Hooke’s law. A spring will stretch 12 cm when a 250-g mass is hung on it. If the 250-g mass is hung on two identical springs, hung sided by side, and both support the 250-g mass, how much will each spring stretch. EXPLAIN.

Answer 1

6 cm

Step-by-step explanation:

Hooke's Law: It states that the strain produced in the body is directly proportional to the stress applied on that body within the elastic limit,

In other words we can say that, the extension produced is directly proportional to the load applied.

Let K be the spring constant.

F = Kx

250 x 981 = K x 12

K = 20437.5 dyne/ cm

Now the two same spings are connected in series.

So new K' = 2 K = 2 x 20437.5 = 40875 dyne / cm

Again use F' = K' x'

250 x 981 = 40875 x'

x' = 6 cm

Thus, each spring stretches by 6 cm.

Answer 2

Hooke's law states that the amount of the stretch in a spring is directly proportional to the amount of force added to it. In the second scenario, there are already two springs and each of them will have to support only half of the downward force of the material attached to it. Therefore, the stretch in each of the spring should only be 6 cm.