Question

1. Assume that hanging a 100 gram mass from a given spring stretches the spring by 2 cm. If two springs of this kind are connected back to back (series configuration), how much would the combination stretch if a 200 gram mass is suspended?

a. 1 cm

b. 2 cm

c. 3 cm

d. 4 cm

e. none of these

2. What would be the answer to the previous question if the two springs were connected in parallel configuration?

a. 1 cm

b. 2 cm

c. 3 cm

d. 4 cm

e. none of these

Answer 1

According to Hook's law, the restoring force on a spring is given by,

where,

F is the force of extension, which is equal to the weight of the body hung (F=mg)

k is the spring constant of the spring

x is the extension produced.

Given,

100g produces an extension of 2cm

Therfore,

m=100g = 0.1kg

F = mg = 0.1x9.8 = 0.98N

x = 2cm = 0.02m

From equation (1)

Hence spring constant k of the spring is 49Nm

a) Two spings of this kind are connected in series

When connected in series, the effective spring constant is given by,

Here both the springs are same, so

Hence in this case,

m = 200g = 0.2kg

F = mg = 0.2x 9.8 = 1.96N

Therefore from equation (1)

Hence the extension produced is 0.08m or 8cm. Since such an option is not available, None of these will be the answer.

b) Two spings of this kind are connected in parallel

When connected in parallel, the effective spring constant is given by,

Here both the springs are same, so

Hence in this case,

m = 200g = 0.2kg

F = mg = 0.2x 9.8 = 1.96N

Therefore from equation (1)

Hence the extension produced is 0.02m or 2cm. So option b is the correct answer.