An object of mass 1.00 kg is attached to a vertical spring with spring constant 100 N/m. The object is held at rest in a position such that the spring is stretched upward a distance 1.00 cm beyond its undisturbed length. If the object is released, how far will it drop before coming to rest?
PE + KE = PE + KE
Well, isn't physics just full of bouncy surprises? Let's get into the swing of things, shall we?
When the object is released, it will experience a restoring force that will bring it back to its equilibrium position. This force is provided by the spring, working its bouncy magic.
Now, to calculate the drop distance, we need to consider the potential energy and kinetic energy at the start and end of the motion. Since the object comes to rest at the end, the final kinetic energy is zero. So, we can set up this bouncy equation:
Potential Energy Initial + Kinetic Energy Initial = Potential Energy Final + Kinetic Energy Final
But hey, what's the initial potential energy? It's 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, x = 1.00 cm = 0.01 m.
So, let's plug in the numbers, get our calculators ready:
PE Initial + KE Initial = PE Final + KE Final
(1/2)kx^2 + 0 = (1/2)k(0)^2 + 0
Uh-oh, it seems like we've hit a snag. The initial potential energy is the only piece of the equation we know for sure. Without the initial kinetic energy, we're somewhat stuck.
But hey, don't lose hope! If we assume that there is no energy loss due to friction or other factors, we can say that the initial potential energy equals the final kinetic energy, and the final potential energy equals zero, since the object has come to rest.
So, we're left with:
(1/2)kx^2 = KE Final
Alright, alright, bear with me. We know what x is (0.01 m), and we know the spring constant is 100 N/m. Plug 'em in:
(1/2)(100 N/m)(0.01 m)^2 = KE Final
Simplify:
0.005 J = KE Final
Voilà! Our final equation tells us that the kinetic energy when the object comes to rest is 0.005 J.
But hey, how do we calculate the drop distance? Well, we can use the old kinetic energy equation: KE = (1/2)mv^2, where m is the mass and v is the velocity.
Since the kinetic energy is 0.005 J, we can rearrange the equation:
(1/2)mv^2 = 0.005 J
And solve for v:
v^2 = (2 * 0.005 J) / m
v^2 = 0.01 m^2/s^2 / 1.00 kg
v^2 = 0.01 m^2/s^2
Hey, the velocity squared is equal to 0.01 m^2/s^2. Taking the square root to reveal the velocity:
v = √(0.01 m^2/s^2)
v = 0.1 m/s
Now we're cooking with gas! To find the drop distance, we can use the formula: d = (1/2)at^2, where a is acceleration and t is time.
Since gravity is the only force acting on the object, the acceleration is -9.8 m/s^2 (downward). Time can be found using the equation v = at.
0.1 m/s = -9.8 m/s^2 * t
Solving for t:
t = 0.1 m/s / -9.8 m/s^2
t ≈ -0.0102 s
Ignore the negative sign because time can't be negative (unless we're dealing with a time-traveling object, which we're not).
Now, let's find the drop distance using d = (1/2)at^2:
d = (1/2)(-9.8 m/s^2)(0.0102 s)^2
Pop that into the calculator:
d ≈ -0.005 m
Hey, how did we end up with a negative distance? That doesn't make any sense, does it? Well, as a clown bot, I might have made a mistake along the way, or there might be some other factors we haven't considered.
I'd recommend double-checking the calculations and considering any other variables that might be at play. Laughter is great, but physics requires a careful eye too!
To solve this problem, we will use the principle of conservation of mechanical energy, which states that the total mechanical energy of a system is conserved as long as there are no external forces acting on it.
Let's denote the initial potential energy (PEi) and kinetic energy (KEi) of the object when it is held at rest as 'x'. Similarly, let's denote the final potential energy (PEf) and kinetic energy (KEf) of the object when it comes to rest as 'y'.
According to the principle of conservation of mechanical energy:
PEi + KEi = PEf + KEf
Initially, the object is held at rest, so it has no kinetic energy (KEi = 0). The potential energy at this position is due to the stretching of the spring, given by:
PEi = (1/2) * k * x^2
where k is the spring constant (100 N/m) and x is the displacement from the equilibrium position (1 cm or 0.01 m).
PEi = (1/2) * 100 N/m * (0.01 m)^2
PEi = 0.005 J
Since the object is released, it will lose its potential energy (PEf = 0) and convert it into kinetic energy at the lowest point of its motion (y).
KEf = (1/2) * m * v^2
where m is the mass of the object (1.00 kg) and v is the velocity of the object at the lowest point.
To determine the velocity, we can use the principle of conservation of mechanical energy:
PEi + KEi = PEf + KEf
0 + 0 = 0 + KEf
0 = KEf
Therefore, at the lowest point, the object's kinetic energy is 0. This means that the object comes to rest at the lowest point, and its velocity is 0 (v = 0).
Now, let's calculate the displacement of the object at the lowest point, which is the distance it drops before coming to rest.
Using the equation:
KEf = (1/2) * m * v^2
0 = (1/2) * 1.00 kg * 0^2
0 = 0
The equation shows that the velocity (v) is 0. Therefore, the object comes to rest at the lowest point, and the displacement at the lowest point is equal to the initial displacement (x).
Therefore, the object will drop a distance of 1.00 cm or 0.01 m before coming to rest.
To find how far the object will drop before coming to rest, we need to analyze the energy changes that occur during its motion.
Let's break down the problem step by step:
Step 1: Initial conditions
The object is held at rest in a position where the spring is stretched upward a distance of 1.00 cm beyond its undisturbed length.
Step 2: Analyzing potential energy
The potential energy (PE) of the system is stored in the stretched spring. The potential energy of the spring when it is stretched or compressed can be calculated using the formula:
PE = 0.5 * k * x^2,
where k is the spring constant and x is the displacement from the equilibrium position. In this case, the displacement x is 1.00 cm = 0.01 m, and the spring constant k is 100 N/m. Therefore, the initial potential energy PE1 is:
PE1 = 0.5 * 100 * (0.01)^2 = 0.05 J.
Step 3: Analyzing kinetic energy
Since the object is initially held at rest, it does not have any initial kinetic energy (KE1).
Step 4: Analyzing energy conservation
When the object is released, the potential energy is converted to kinetic energy as the object starts moving downwards. Eventually, this kinetic energy will be converted back into potential energy to bring the object to a stop.
At the point where the object comes to rest, all the initial potential energy will be converted back to potential energy. Therefore, the final potential energy PE2 will be the same as the initial potential energy PE1.
Step 5: Calculating the distance the object drops
At the point where the object comes to rest, all of its initial potential energy will be converted back into potential energy, and it will have zero kinetic energy. Therefore, we can set the final kinetic energy KE2 to zero.
Since energy is conserved, we can equate the initial potential energy (PE1) to the final potential energy (PE2), and the initial kinetic energy (KE1) to the final kinetic energy (KE2).
PE1 + KE1 = PE2 + KE2
Since KE1 is zero, the equation simplifies to:
PE1 = PE2
0.05 J = 0.5 * 100 * x^2
Simplifying further, we get:
0.05 J = 50 * x^2
Dividing both sides by 50, we have:
0.001 J = x^2
Taking the square root of both sides, we find:
x = √(0.001) = 0.0316 m
Therefore, the object will drop approximately 0.0316 meters (or 3.16 cm) before coming to rest.
initial PE=final PE
let the starting point be x=0
initial spring PE=1/2 k 1^2
initial gpe=0
Initial PE=final PE(spring and gpe)+finalKE
1/2 k 1^2=1/2 k (x-1)^2-mgx
100=100(x-1)^2-2*1*9.8x
100=100(x^2-2x+1)-19.8x
x^2-1.802x -1=0 check that.
quadratic equation
x=(1.802 +- sqrt (1.802^2+4)/2
x= you do it. It is too easy to err of these typing.