If a 0.1 kg mass is suspended at rest from a spring near the Earth's surface, the distance that the spring is stretched is measured to be 1.0 cm. What is the spring constant of the spring (remember the MKS units)?

A mass of 2 kg is attached to a spring with constant 18 N/m. How far is the spring stretched?

A spring with a spring constant of 28.9 N/m is attached to different masses, and is stretched 5cm. What is the force exerted on the spring?

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To find the spring constant, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position. The equation for Hooke's Law is F = kx, where F is the force, k is the spring constant, and x is the displacement.

For the first question, we are given the mass (m = 0.1 kg) and the displacement (x = 1.0 cm = 0.01 m). Since the mass is in equilibrium (at rest), the force exerted by the spring is balanced by the force of gravity acting on the mass. Therefore, we can equate the two forces: mg = kx.

Rearranging the equation, we can solve for the spring constant:

k = mg / x
k = 0.1 kg * 9.8 m/s^2 / 0.01 m
k = 98 N/m

So, the spring constant of the spring in the first scenario is 98 N/m.

For the second question, we are given the mass (m = 2 kg) and the spring constant (k = 18 N/m). We need to find the displacement of the spring (x).

Using Hooke's Law, we can rearrange the equation to solve for x:

F = kx
x = F / k
x = mg / k
x = 2 kg * 9.8 m/s^2 / 18 N/m
x = 1.08 m

Therefore, the spring is stretched by 1.08 meters.

Lastly, for the third question, we are given the spring constant (k = 28.9 N/m) and the displacement (x = 5 cm = 0.05 m). We need to find the force exerted on the spring (F).

Using Hooke's Law: F = kx
F = 28.9 N/m * 0.05 m
F = 1.45 N

So, the force exerted on the spring is 1.45 Newtons.