Question

A 2 kg mass hangs motionless from a partially stretched spring having a spring constant of 172 N/m. What is the magnitude of the force (in N) that would be required to pull the mass down by an additional 5 cm? Never include units with your answer.

Answer 1

Answer:

8.608 N

Explanation:

m = 2 kg, K = 172 N/m

Let the extension is x.

F = K x

x = mg / K = 2 x 9.8 / 172 = 0.114 m

now d = 5 cm = 0.05 m

Let additional force is F.

(mg + F) = K (x + d)

(2 x 9.8 + F) = 172(0.114 + 0.05)

19.6 + F = 28.208

F = 8.608 N

Answer 2

The magnitude of the force (in N) that would be required to pull the mass down by an additional 5 cm is 8.6 N.

Force applied to the elastic material

The magnitude of the force (in N) that would be required to pull the mass down by an additional 5 cm is determined by using Hooke's law as shown below;

F = kx

where;

• k is the spring constant
• x is the extension of the material

F = 172 x (0.05)

F = 8.6 N

Thus, the magnitude of the force (in N) that would be required to pull the mass down by an additional 5 cm is 8.6 N.

Learn more about Hooke's law here: https://brainly.com/question/2648431