A 0.20-kg mass is hung from a vertical spring of force constant 55 N/m. When the

spring is released from its unstretched equilibrium position, the mass is allowed to
fall. Use the law of conservation of energy to determine
(a) the speed of the mass after it falls 1.5 cm
(b) the distance the mass will fall before reversing direction

The weight is

F = M g = 1.96 N
The equilibrium deflection is
Xe = F/k = 3.56*10^-2 m = 3.56 cm.
It will fall to twice that deflection.

Spring and gravitational potential energy, and kinetic energy, are zero when it is initially dropped

(a) The initial total energy, relative to the initial position, is zero.

After falling x = 0.015 m,
GravPE + SpringPE + KE = 0
-M g x + (1/2)kx^2 + (M/2) V^2 = 0

Solve for V.

(b) For max deflection, compute the other location where KE = 0

(1/2)kX^2 = M g X
X = 2 Mg/k, twice the equilibrium deflection

Well, well, well, looks like we have a physics problem on our hands! Let's dive right in and have some fun with conservation of energy.

(a) To determine the speed of the mass after it falls 1.5 cm, we can start by calculating the potential energy stored in the spring when it's stretched. The potential energy (PE) of a spring is given by the equation PE = (1/2)kx^2, where k is the force constant of the spring and x is the distance stretched.

In this case, the force constant is 55 N/m and the distance stretched is 0.015 m (converting 1.5 cm to meters), so let's plug those values in:

PE = (1/2)(55 N/m)(0.015 m)^2

Simplifying that little equation will give us the potential energy stored in the spring.

Next, we can equate the potential energy to the kinetic energy (KE) of the mass when it falls. That's because energy is conserved, baby! The kinetic energy is given by the equation KE = (1/2)mv^2, where m is the mass and v is the velocity.

So we set the potential energy equal to the kinetic energy and get:

(1/2)mv^2 = (1/2)(55 N/m)(0.015 m)^2

Now it's just a matter of algebraic manipulation to solve for v. Do the math and you'll have your answer to part (a)!

(b) Ahhh, the distance the mass will fall before reversing direction. Well, remember that potential energy we calculated earlier? Yeah, we're going to use it again!

When the mass has fallen, it will reach its maximum velocity. At this point, all potential energy has been converted into kinetic energy. The important thing to remember here is that any extra energy will go into the kinetic energy of the mass. So, when the mass falls, the potential energy will be converted to kinetic energy until all potential energy is exhausted.

Once all the potential energy is gone, the mass will reach its highest point, and that's when it turns around and starts to fall back. At this point, all its energy is in the form of kinetic energy, and its velocity will decrease as it moves upward.

So, to determine the distance the mass will fall before reversing direction, we need to find the maximum potential energy of the spring. We already calculated that earlier, remember? Now, just divide that potential energy by the mass of the object and the force of gravity, and voila! You'll have your answer to part (b)!

Remember, physics doesn't have to be boring. Embrace the humor and enjoy solving those problems!

To determine the speed of the mass after it falls 1.5 cm, we can use the law of conservation of energy. The total mechanical energy of the system is conserved, so we can equate the initial potential energy with the final kinetic energy.

(a) Speed of the mass after falling 1.5 cm:

1. Calculate the potential energy of the mass at its initial position:

Potential Energy (PE) = m * g * h, where m is the mass, g is the acceleration due to gravity (9.8 m/s^2), and h is the height.

PE = (0.20 kg) * (9.8 m/s^2) * (0.015 m)
PE = 0.0294 J

2. Calculate the potential energy at the final position (the bottom of the fall):

When the mass reaches the bottom, all of its potential energy is converted into kinetic energy, so:

Potential Energy (PE) = 0 J

3. Equate the initial potential energy with the final kinetic energy:

Initial Potential Energy = Final Kinetic Energy

0.0294 J = (1/2) * m * v^2, where v is the final velocity.

4. Solve for v:

v^2 = (2 * 0.0294 J) / 0.20 kg
v^2 = 0.294 J / 0.20 kg
v^2 = 1.47 m^2/s^2

v = √(1.47 m^2/s^2)
v ≈ 1.21 m/s

Therefore, the speed of the mass after falling 1.5 cm is approximately 1.21 m/s.

(b) The distance the mass will fall before reversing direction:

At the highest point of the mass's motion, it will momentarily come to rest before reversing direction. Therefore, the initial potential energy will be equal to the total mechanical energy at this point.

1. Calculate the initial potential energy (at the highest point):

Potential Energy (PE) = m * g * h

PE = (0.20 kg) * (9.8 m/s^2) * (h)
PE = 1.96 h J

2. Calculate the maximum height the mass will reach:

At the highest point, the kinetic energy is zero, so the potential energy is equal to the initial potential energy:

1.96 h J = 0.0294 J

Solve for h:

h = 0.0294 J / 1.96 J
h ≈ 0.015 m

So, the distance the mass will fall before reversing direction is approximately 0.015 m.

To determine the speed of the mass after it falls 1.5 cm and the distance the mass will fall before reversing direction, we can use the law of conservation of energy.

The law of conservation of energy states that the total mechanical energy of a system remains constant if there are no external forces acting on it. In this case, the only force acting on the system is the force of gravity.

Let's break down the problem step by step:

(a) Determining the speed of the mass after it falls 1.5 cm:
1. First, we need to calculate the potential energy (PE) stored in the spring when it is stretched by 1.5 cm.
The potential energy of a spring is given by the formula: PE = (1/2)kx^2, where k is the force constant of the spring and x is the displacement from its equilibrium.

Plugging in the values, we have: PE = (1/2)(55 N/m)(0.015 m)^2 = 0.012375 J

2. According to the law of conservation of energy, this potential energy is then converted into kinetic energy (KE) when the mass falls.
The kinetic energy of an object is given by the formula: KE = (1/2)mv^2, where m is the mass of the object and v is its velocity.

We can equate the potential energy to the kinetic energy: PE = KE

In this case, PE = 0.012375 J, and we are trying to find v.

Therefore, 0.012375 J = (1/2)(0.20 kg)(v^2)

Solving for v, we get: v^2 = (0.012375 J) * 2 / (0.20 kg)

Calculating the value, we find: v^2 = 0.12375 J/kg

Taking the square root, we get: v ≈ 0.351 m/s

So, the speed of the mass after it falls 1.5 cm is approximately 0.351 m/s.

(b) Determining the distance the mass will fall before reversing direction (maximum displacement):
1. The maximum displacement occurs when all the potential energy is converted into kinetic energy and the mass momentarily comes to rest.

Setting the potential energy (PE) equal to zero (when the mass comes to rest), we have: PE = (1/2)kx^2 = 0

Rearranging the equation, we find: x^2 = 0 / (1/2)k = 0

Therefore, x = 0. This means that the maximum displacement is 0, which indicates that the mass does not reverse direction.

In conclusion:
(a) The speed of the mass after it falls 1.5 cm is approximately 0.351 m/s.
(b) The mass will not reverse direction and will not fall any further than the initial displacement of 1.5 cm.

ETi=ETf

Ek+Eg+Ee =Ek+Eg+Ee
Intial/final speed is zero, and having a ref point from the bottom, hf is zero and xi will be zero since in the beging the spring is at equiibrim. Finally ur hi=xf, since the strech will be the distaince traveled.
From this info and the equation we can get:

Eg+Ee= Eg+Ee
mghi+0.5(k)(xi)^2 = mghf+0.5(k)(xf)^2
mghi=0.5(k)(hi)^2
mg=0.5(k)(hi)
mg/0.5(k)=hi
sub in values to get
(0.20)(9.80)/(0.5)(55)=hi
0.071 meters= hi