Question

A 2 kg mass is hung from a vertical spring and gently lowered to its equilibrium position, which stretches the spring 10 cm. The spring is returned to its unstretched position, the 2 kg mass is replaced with a 3 kg mass, and the mass is released (not gently lowered). How far will the spring stretch before the mass stops and bounces back up?

Answer 1

The spring will stretch by 0.16 m before the mass stops and bounces back.

Step-by-step explanation:

When a 2 kg mass is hung from a vertical spring and gently lowered to its equilibrium position, it stretches the spring by 10 cm. The equation for the force exerted by a spring is F = kx, where F is the force, k is the force constant, and x is the displacement of the spring from its equilibrium position. We can use this equation to find the force constant of the spring.

Given that the displacement of the spring is 10 cm, we convert it to meters by dividing it by 100: 10 cm / 100 = 0.1 m.

Using the equation F = kx, we can find the force constant: k = F / x = (mg) / x = (2 kg × 9.8 m/s²) / 0.1 m = 19.6 N/m.

Now, let's consider the 3 kg mass. Since it is released instead of gently lowered, it experiences an additional force due to gravity. The force exerted by the spring will be equal to the weight of the mass plus the force exerted by the spring.

Using the same equation F = kx, we can find the displacement of the spring: x = F / k = (mg + kx) / k = (3 kg × 9.8 m/s² + 19.6 N/m × 0.1 m) / 19.6 N/m = 0.16 m.

Therefore, the spring will stretch by 0.16 m before the mass stops and bounces back.

Answer 2

When a 3 kg mass is released on a spring with a spring constant of 19.6 N/m, the spring will stretch 14.7 cm.

Step-by-step explanation:

To determine how far the spring will stretch when a 3 kg mass is released, we can use Hooke's Law. Hooke's Law states that the force exerted by a spring is proportional to the displacement from equilibrium. The formula is

F = -kx,

where F is the force, k is the spring constant, and x is the displacement. In this case, the spring constant k is not given, but we can find it using the information provided.

First, we know that a 2 kg mass stretches the spring by 10 cm (0.1 m) before reaching equilibrium. We can use this information to find the spring constant.

Using Hooke's Law, we can set up the equation F = -kx and solve for k. The force F is the weight of the object, which is mg, where m is the mass and g is the acceleration due to gravity.

Substituting the values, we get

-k(0.1) = -2(9.8).

Simplifying the equation, we find that k = 19.6 N/m.

Now we can use this spring constant to determine the displacement of the spring when a 3 kg mass is released.

Using Hooke's Law, we can set up the equation F = -kx and solve for x.

The force F is the weight of the object, which is mg.

Substituting the values, we get

-19.6x = -3(9.8).

Solving for x, we find that x = 0.147 m, or 14.7 cm.

Therefore, the spring will stretch 14.7 cm when the 3 kg mass is released and bounces back up.