My answer was 19.8 cm but it was incorrect.

A basket of negligible weight hangs from a vertical spring scale of force constant 1500 N/m.

If you suddenly put a 3.00 kg adobe brick in the basket, find the maximum distance that the spring will stretch. Answer: 3.92 cm

If, instead, you release the brick from 1.00 m above the basket, by how much will the spring stretch at its maximum elongation?

My answer was 19.8 cm but it was incorrect.

At equilibrium, the spring stretches an amount X given by

k X = W = M g
X = 3.00*9.81/1500 = 1.96 *10^-2 m = 1.96 cm
When the brick is dropped suddenly, it stretches twice as far and oscillates about the 1.96 cm deflection equilibrium position.
The maximum deflection is 2*1.96 = 3.92 cm. I assume you got that part right

In the second case, use conservation of energy. The brick falls a distance 1.00 + Xmax and converts gravitational P.E. = M g (1 + Xmax) into spring energy (1/2)k Xmax^2
29.4 (1 + Xmax) = 750 Xmax^2
Solve the quadratic equation.

Xmax = (1/1500)[29.4 + sqrt(864+88200)] = 0.219 m = 21.9 cm

You may have forgotten to include the additional PE loss during stretching

Well, well, well, looks like you really stretched it with that answer! But hey, no worries, we'll figure this out together.

Now, imagine that gravity is like that one friend who always brings you down. When you release the brick from 1.00 m above the basket, it's going to experience quite the drop. And just like my enthusiasm when I see cotton candy, that brick is going to gain some speed!

But hold on, the spring scale is there to catch the brick in all its majestic falling glory. That means it's going to stretch and elongate, just like a rubber band on steroids!

Now, to find out how much that spring will stretch, we need a bit of physics in the mix. Specifically, we need to consider the potential energy of the brick when it's in the air, which will then be converted into the potential energy stored in the stretched spring.

So, let's do the math. The potential energy of the brick can be calculated using the formula PE = mgh, where m is the mass of the brick, g is the acceleration due to gravity (about 9.8 m/s²), and h is the height from which the brick is released.

In this case, we have m = 3.00 kg and h = 1.00 m. So, the potential energy is 3.00 kg × 9.8 m/s² × 1.00 m = 29.4 J.

Now, remember that potential energy is also equal to the spring potential energy, which can be calculated using the formula PE_spring = (1/2)kx², where k is the force constant of the spring and x is the stretch or elongation of the spring.

In this case, we have k = 1500 N/m, and we need to find x. Let's go step by step and solve for x.

29.4 J = (1/2) × 1500 N/m × x²

Now, we divide both sides of the equation by (1/2) × 1500 N/m to isolate x².

(29.4 J) / [(1/2) × 1500 N/m] = x²

x² = (29.4 J) / [(1/2) × 1500 N/m]

x² = 0.0392 m

Taking the square root of both sides, we find:

x = √(0.0392 m) = 0.198 m

Now, hold your horses before you start celebrating! We need the answer in cm, so let's convert that to centimeters.

0.198 m × 100 cm/m = 19.8 cm

And there you go! The maximum elongation of the spring when the brick is released from 1.00 m above the basket is 19.8 cm.

Oh wait, my red nose is itching again! Looks like I made a mistake earlier. That answer is wrong! My sincerest apologies for clowning around with your answer.

The correct answer is 0.198 m, or if we're feeling fancy, 19.8 cm. I guess it's time to bring out the flower that squirts water and admit my mistakes. I hope that clears things up!

To find the maximum elongation of the spring when the brick is released from 1.00 m above the basket, you can use the principle of conservation of mechanical energy.

The potential energy of the brick when it is released is given by the equation PE = mgh, where m is the mass of the brick, g is the acceleration due to gravity, and h is the height from which the brick is released.

The potential energy is converted into the elastic potential energy of the spring when the brick reaches its maximum elongation.

The elastic potential energy of a spring can be given by the equation PE = (1/2)kx^2, where k is the force constant of the spring and x is the elongation of the spring.

Since the potential energy of the brick is equal to the elastic potential energy of the spring at its maximum elongation, we can equate the two equations:

mgh = (1/2)kx^2

Rearranging the equation, we get:

x = √((2mgh)/k)

Plugging in the values, we have:

m = 3.00 kg
g = 9.8 m/s^2
h = 1.00 m
k = 1500 N/m

x = √((2 * 3.00 kg * 9.8 m/s^2 * 1.00 m) / 1500 N/m)

Calculating the value, we get:

x = √(58.8/1500)

x ≈ 0.169 m

Converting the value to centimeters, we get:

x ≈ 16.9 cm

Therefore, the maximum elongation of the spring when the brick is released from 1.00 m above the basket is approximately 16.9 cm.

To find the maximum distance that the spring will stretch when the brick is released from a height of 1.00 m above the basket, we need to consider the potential energy that is converted into the spring potential energy.

First, let's determine the potential energy of the brick when it is released from a height of 1.00 m above the basket. The potential energy (PE) of an object at a certain height is given by the equation:

PE = m * g * h

where m is the mass of the object, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height.

In this case, the mass of the brick is 3.00 kg and the height is 1.00 m. Thus, the potential energy of the brick is:

PE = 3.00 kg * 9.8 m/s^2 * 1.00 m
= 29.4 J

Now, since energy is conserved, this potential energy will be converted into the potential energy of the spring at its maximum elongation. The potential energy of a spring can be calculated using the formula:

PE = (1/2) * k * x^2

where k is the force constant of the spring and x is the displacement from equilibrium.

In this case, the force constant of the spring is given as 1500 N/m. We need to find the displacement at the maximum elongation, which we'll call x_max.

Rearranging the formula and solving for x_max:

2 * PE = k * x_max^2
x_max^2 = (2 * PE) / k
x_max = √((2 * 29.4 J) / 1500 N/m)
x_max = 0.092 m = 9.2 cm

Therefore, the maximum distance that the spring will stretch when the brick is released from a height of 1.00 m above the basket is 9.2 cm.