A .20 kg ball is attached to a vertical spring. The spring constant is 28 N/m. The ball is supported initially so that the spring is neither stretched or compressed, is released from rest. How far does the ball fall before brought to a momentary stop by the spring?

Our teacher gave us a hint:
PE+KE=PE+KE

But I think that there is something missing, perhaps subscripts (initial and final). And if that's the case, then, I'll get PE=PE.

Also, someone else in class explained that PE=mgh and solved the problem, but I do not know how she got .5 for h.

Please help, so confused. Thank you.

OK: The ball has gravity pulling on, while is supported by something. When that support is moved, the spring holds the ball. The force on the spring is mg, and it gains velocity as it falls, thus gaining KE.

assume the ball falls h before stopping.
The energy then in the spring is 1/2 k h^2, and it is equal to the gravity PE that Earth released (mgh).

mgh=1/2 k h^2
mg= 1/2 k h

h= 2mg/k = 2*0.2*9.8/28 m= you do it.

In my head, I see it about .14m, not even close to .5m

Ah, physics problems can be quite a spring-loaded challenge! Let's unravel this and give it a humorous twist.

First, let's address the formula that your teacher hinted at: PE + KE = PE + KE. It's like a secret code that doesn't reveal much. But fear not, let's decode it together.

PE refers to potential energy, which we can think of as the energy stored in the spring when it's compressed or stretched. KE stands for kinetic energy, which is the energy of movement.

When the ball is at its initial position, it has no kinetic energy because it's at rest. However, it does have potential energy since it's attached to the spring, which might be compressed or stretched.

When the ball is momentarily stopped by the spring, it will have no kinetic energy again because it's at rest. However, this time the spring will have potential energy stored in it.

So using the hint, we can say:

Potential energy before = Potential energy after

Now, let's address the "missing" subscripts. You're absolutely right, we need to distinguish between the initial and final positions. Let's label them as "initial" and "final" respectively:

Potential energy (initial) + Kinetic energy (initial) = Potential energy (final) + Kinetic energy (final)

Since the ball is initially at rest, the initial kinetic energy is zero. The final kinetic energy is also zero because the ball comes to a momentary stop.

Now, let's focus on the potential energy. The potential energy of an object in a gravitational field is given by PE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height.

In this case, we don't have a height, but we have a displacement x of the ball caused by the spring. The potential energy stored in the spring is given by PE = (1/2)kx^2, where k is the spring constant.

The distance the ball falls before being brought to a momentary stop is the displacement caused by the spring. So, using the equation, (1/2)kx^2 = mgh, we can solve for x.

But wait, why is h equal to 0.5? That seems a bit mysterious, doesn't it? Well, it seems your classmate was pulling a bit of a comedy trick there. The value of h should indeed be zero since the ball doesn't fall in a gravitational field, but instead, it moves due to the spring. So h = 0, not 0.5.

Now that the comedic confusion is cleared up, we can solve the problem using the correct equation. Set (1/2)kx^2 = mgh, substitute the given values, solve for x, and voila! You'll find the distance the ball falls before being stopped by the spring.

Remember, learning can be like juggling, sometimes a bit tricky but always entertaining. Keep up the good work!

To solve this problem, you can use the law of conservation of mechanical energy. The total mechanical energy of the system (ball and spring) remains constant throughout the motion.

Let's break down the problem step-by-step:

Step 1: Find the initial potential energy (PE) of the ball.
The initial potential energy of the ball is given by the formula PE = mgh, where m is the mass of the ball, g is the acceleration due to gravity, and h is the height from which the ball is released. However, in this case, the ball is initially supported by the spring, so it is not at a height h. Therefore, the initial potential energy is zero (PE_initial = 0).

Step 2: Find the initial kinetic energy (KE) of the ball.
Since the ball is initially at rest, the initial kinetic energy is zero (KE_initial = 0).

Step 3: Find the final potential energy (PE) of the ball.
When the ball is momentarily brought to a stop by the spring, all its initial potential energy (zero) is converted into potential energy stored in the spring. The formula for the potential energy stored in a spring is PE = (1/2)kx^2, where k is the spring constant and x is the displacement of the spring from its equilibrium position.

Step 4: Find the final kinetic energy (KE) of the ball.
When the ball reaches its maximum displacement from equilibrium, its velocity is zero. Therefore, the final kinetic energy is also zero (KE_final = 0).

Step 5: Equate the total initial mechanical energy to the total final mechanical energy.
Using the law of conservation of mechanical energy, we can write the equation as:
PE_initial + KE_initial = PE_final + KE_final

Step 6: Calculate the displacement (x) of the ball from its equilibrium position.
Since all the potential energy is converted into the potential energy stored in the spring, we can rewrite the equation as:
0 + 0 = (1/2)kx^2 + 0

Now, let's calculate the displacement:

(1/2)kx^2 = 0
(1/2)(28 N/m)x^2 = 0

Since we have (1/2) multiplied by a constant, the whole term becomes zero:
x^2 = 0

Therefore, the displacement (x) of the ball from its equilibrium position is zero. This means that the ball does not fall at all before it is brought to a momentary stop by the spring.

To solve this problem, you can use the principle of conservation of mechanical energy. The hint from your teacher, "PE + KE = PE + KE," is correct, but it does require specifying initial and final states for potential energy (PE) and kinetic energy (KE).

First, let's establish the initial and final states. The initial state is when the ball is supported so that the spring is neither stretched nor compressed. The final state is when the spring brings the ball to a momentary stop.

Let's assign subscripts to the initial and final states:
Initial state: i
Final state: f

Now we can rewrite the conservation of mechanical energy equation as follows:
PEi + KEi = PEf + KEf

In the initial state (PEi + KEi), the ball is at rest, so the initial kinetic energy (KEi) is zero. The initial potential energy (PEi) is also zero because the spring is neither stretched nor compressed.

Therefore, we can simplify the equation to:
KEi = PEf + KEf

Since the spring is what affects the potential energy (PE), we can rewrite the equation using the potential energy formula:
0 = mghf + KEf

Now, let's address the value of "h." In this context, "h" represents the change in height. When the ball falls, it loses height, so we need to find the difference between the initial and final heights. In this case, since the ball is supported initially, it has not fallen yet, so the initial height (hi) is zero. The final height (hf) is the distance the ball falls before being brought to a momentary stop by the spring.

Now we can modify the equation:
0 = (mg * hf) + KEf

We can then rearrange the equation to solve for hf:
hf = -KEf / mg

To find KEf, we need to determine the final kinetic energy. At the moment the ball is brought to a stop by the spring, all of its initial potential energy (PEi) will be converted into kinetic energy (KEf). Therefore, we can substitute PEi for KEf:
hf = -PEi / mg

Now, let's find PEi:
The potential energy stored in a spring is given by the formula: PE = (1/2)kx^2, where k is the spring constant and x is the displacement from the equilibrium position.

In this case, the spring is neither stretched nor compressed at the initial state, so x = 0. Therefore, PEi = 0.

Substituting PEi = 0 into the equation, we get:
hf = -0 / mg

Finally, we can conclude that the distance the ball falls before coming to a momentary stop by the spring is zero.

In summary, the ball does not fall at all (hf = 0) before being brought to a momentary stop by the spring.