In physics, the kinetic energy of an object is the energy that it possesses due to its motion.
It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes. The same amount of work is done by the body when decelerating from its current speed to a state of rest.
In classical mechanics, the kinetic energy of a non-rotating object of mass m traveling at a speed v is
1
2
m
v
2
{\displaystyle {\begin{smallmatrix}{\frac {1}{2}}mv^{2}\end{smallmatrix}}}
. In relativistic mechanics, this is a good approximation only when v is much less than the speed of light.
The standard unit of kinetic energy is the joule, while the English unit of kinetic energy is the foot-pound.
Source: Shankar Yale OCW physics
I have three questions here:
1. K_avg is 3/2kT, sure. But isn't this the kinetic energy of one particle only? So why isn't the answer multiplied by avogadro's number (because one mole).
2. When doing the "typical velocity" derivation, I noticed that they used...
Kinetic energy = 1/2 m V^2
I was thinking about this and thought another formula…
E = mc^2
These look very similar except for the multiplication by 1/2.
Let’s say you take a kilogram ball of uranium and accelerate it to the speed of light. I know, I know. You can’t. But let’s say you did...
FOr this,
Use alternate coordinate system
With ##ȳ##-axis parallel to incline and ##x̄##-axis parallel to the x-axis. Kinetic energy using this alternate coordinate system is ##T = \frac{1}{2}M\dot x_p^2 + \frac{1}{2}mR^2\dot \phi^2 + \frac{1}{2}m(\dot x̄^2 + \dot ȳ^2 + 2\dot x̄ \dot ȳ...
I expanded ET1=ET2 to get
(Total energy at top) 1/2mv^2+mgh = 1/2kx^2 (Total energy at bottom)
Rearanged i got
k = (mv^2+2mgh)/x^2
so [(73)(20)^2+2(73)(9.8)(52)]/0.465^2
=479137.945N/m
I did some algebra and got the final equation mgh=1/2MR^2omega^2+1/2(2/5MR^2)omega^2, then plugged in the numbers and got an omega value of 111.09. Using 1.7 as h instead of cos(17)*1.7=1.63 I got 114.09. Both answers were wrong. Am I missing something, or did I just screw up the math?
part i)
i did 1/2 * 1700 * v^2
i dont know what v is...
so how do i solve it?
part ii)
i calculated it correctly by 440*25
please explain in detail why i used 440?
and part d)
i did 1.7*10^4 = 48000/t
my t= 2.82 s
but correct answer is 3.5s
Does the high temperature increase the kinetic energy of molecules or atoms, or does the high kinetic energy of atoms or molecules increase the temperature ?
I'm so curious about this. Which concept is more accurate between the two
Homework Statement: Real world application of freshman physics
Relevant Equations: TBD
This is not a homework question, this is relevant to my work. It seems simple enough (introductory) but I keep running into problems.
An electron is emitted from an surface (material is irrelevant, could...
I calculated the net force. I got 6500. I determined this is bigger than the force of static friction so the force of friction acting on the object must be kinetic. From there I got kinda lost. I know for the component to not slip Fnetx = 0 and Fnety = 0. But I'm not sure what to do from there...
Hello everyone,
I'm currently working on a physics problem involving the rotation of a 5 kilogram ##M=5## solid sphere subjected to a force of 5 newtons ##F=5##, and I've encountered an inconsistency in my calculations. I'm seeking guidance or insights into where I might have gone wrong.
My...
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I am working on this problem and I think I almost solved it, but then I noticed, that I do not know what values I have for dn, n and dθ.
Can anyone help me with this?
Suppose in a different star system, a space shuttle sized spacecraft acquired a solar mass energy equivalent amount of kinetic energy, then passed through our solar system. While it was passing through the solar system would the craft’s gravitational effects be more similar to the space shuttle...
Let's consider a reaction A (reactant) -> B(product) and activated complex is denoted by C.
This graph ( potential energy vs reaction coordinate ) tells us that reactant need some amount of activation energy (Ea) to convert in product, which has low potential energy which is shown here in...
For this,
From the work kinetic energy theorem, if we assume that the book and the earth is the system, and that the finial and inital speed of the system is zero, then is the work KE theorem there is no net work done on the system. However, clearly there is work done on the system is shown by...
Given that there is a cylinder rolling without slipping down an incline, the method I was taught to represent the KE of the cylinder was:
##KE_{total} = KE_{translational} + KE_{rotational}##
##KE_{total} = \frac {1} {2} mv_{cm}^2 + \frac1 2 I \omega^2## Where "cm" is the center of mass, and...
The cylinder in question would have a moment of inertia of ~1.67kg*m² and rotational KE of 2.058J. At the point of impact also, assuming the body hits the sphere at a 90deg angle after traversing 90deg of displacement, it should(?) exert a force of 1.31N - enough to give an acceleration of...
Hello,
I've seen in a few books on solid state physics that one can deduce an expression for average K.E.:
$$<\:K.E.>\:=E_c+3/2\:k_B\:T$$
from the following:
$$<\:K.E.>\:=\:\frac{\int \:\left(E-E_c\right)g\left(E\right)f\left(E\right)dE}{\int \:g\left(E\right)f\left(E\right)dE}$$
I can't...
I'm watching a video about " What is a black body?". That video said when the light interacts with the surface of a body, the electron and proton start oscillating. The electrons gain more transferred energy from the light that became its kinetic energy, rather than the proton because its mass...
I could not find any derivations in the litterature, except for the expected value of the energy flux expression itself:
$$\overline{\Phi_{effusion,\epsilon}} = \overline{\dot{N_{ef}}}\overline{\epsilon_{ef}}=\frac{3Nl}{2A}\sqrt{\frac{(k_BT)^3}{2\pi m}}$$
I've started off by calculating the...
I would guess that by multiplying the pressure exerted by the shockwave on the body, and then the resulting force - here ~69 Newtons - per the distance the shockwave passed through when traversing body A, I could get the work done but I’m not sure if it’s that easy and whether or not I should...
W_ext is the external work done on B and C, which is 12 J
Delta K_tot is the internal work, which is the work done by A on B plus the work done by A on C
Delta K_tot = 5
Solving for \Delta U, we find that the change in potential energy is 7 J
This answer says otherwise...
Dear Forum,
I am solving for the expectation value of the kinetic energy for the deuteron (Krane problem 4.3). I must be missing something since this has become far more complicated than I remember.
The problem is as follows:
## <T> = \frac{\hbar^{2}}{2m} \int_{0}^{\infty}...
I am seeing conflicting definitions of degree of freedom in my textbook. If I look at the definition given as per screenshot below then it is the number of independent terms/variables/coordinates used to define the energy of a molecule. But, if I look at the statement of Equipartition of energy...
First I found work:
W=(3.85x10^5)(2.45x10^8)
W= 9.43x10^13
Then used that for difference of kinetic energy:
9.43x10^13 = (1/2) (4.55x10^4)v2^2 - (1/2)(4.55x10^4)(1.22x10^4)^2
9.43x10^13 = (22750)v2^2 - 3.386x10^12
9.43x10^13 + 3.386x10^12 = (22750)v2^2
9.77x10^13 = 22750v2^2
9.77x10^13/22750...
I'm a little confused because my teacher used Bill's 500J of work for the kinetic energy equation and I don't understand why. I used the net work, so 300J, to find the speed and I'm not sure why that's wrong. Wouldn't friction make the wagon move slower than if there was no friction? So why...
hello guys, I wanted to ask whether I can just consider/think about this as being rotation around a fixed axis in a plane representing it as if it was 'just' a rod. This is mainly so that for the kinetic energy in the second position is where if we think about it in just a plane. Is this...
Kindly help me solve this question. The only thing so far that I know in this question is that energy is conserved and the momentum of Alpha particle will equal momentum of Thorium.
Spacetime expands at an accelerated rate and the particles with movement associated to this expansion are coupled to the Hubble flow. In many papers that I've read, objects coupled to the Hubble flow are treated as if they have some velocity and kinetic energy associated with it.However, can...
Young & Freedman 13th ed, Exercise 7.81
Starting with the crate, here is its free-body diagram:
In accordance with Newton's First Law:
$$ \Sigma F_y = 0 = n+(-w_c \cos{\alpha}) $$
Thus ## n = w_c \cos{\alpha} ##.
And according to Newton's Second Law:
$$ \Sigma F_x = m_c a_x = w_c...
Since there is no friction : $$ m \ddot{x} = 0 $$ (no x motion).For the kinetic energy , I've tried: $$ K = 1/2 I_{cm} \dot{\alpha}^2 + 1/2 m v^2_cm = 1/2 I_{cm} \dot{\alpha}^2 + 1/2 m \dot{z}^2$$ . Giving me a weird expression , shouldn't the kinetic energy just be half the the moment...
I tried just using the formula for kinetic energy but that was apparently the wrong answer. The answer key says it's (1/6)mv². I don't understand how they got that answer. Could someone explain?
I am trying to reproduce the results from this paper. On page 10 of the paper, they have an equation:
$$ \frac{S}{T}=\int dt\sum _{n=0,1} (\dot{c_n}{}^2-c_n^2 \omega _n^2)+11.3 c_0^3+21.5 c_0 c_1^2+10.7 c_0 \dot{c_0}{}^2+3.32 c_0 \dot{c_1}{}^2+6.64 \dot{c_0} c_1 \dot{c_1} \tag{B12} $$
where they...
If you want to find out the kinetic energy (the unit is GJ as in other cannons) of the gunpowder cannon that appears in Monster Hunter, what do you need and how can you calculate it?
I am wondering if it is possible to calculate either the Kinetic Energy or Rotational Kinetic Energy of an object if we have the Power (kW), Torque (Nm), and Speed (RPM) of the object.
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This is my answer:
$$KE_{total}=KE_{centermass}+KE_{uppermass}+KE_{bottommass}$$
$$KE_{total} = \frac 1 2 (mv^2 + 2m(\vec {v} + \vec {wL})^2) $$
But, the solution manual says that the answer is this:
$$KE_{total} = \frac 1 2 (mv^2 + 2m(v^2+w^2L^2)) $$
I think he regard this composite body as...
It is usual, in plasma physics, to combine the kinetic description for electron fluid and the hydrodynamic description for ion fluid, when studying the plasma properties or the dynamics of the electrostatic waves.
I am wondering what are the physical meaning and limits of such an approach.
Thanks.
A point charge of value q=8uC is released from rest at a point 1.5m away from the center of the axis of a ring with uniform charge density 3uC/m. The ring has a radius of 10 cm. What is the kinetic energy of this charge when it is 4.5 cm from the center of the charge ring, considering that it is...
I am planning to teach a school astronomy group about energy. Most people seem to accept that there are two types:
kinetic energy, resulting from movement;
potential energy, resulting from position in a force field with a potential gradient (convertible to KE if the object is allowed to move...
I'm not interested in the mathematical derivation, the mathematical derivation already is based on the assumption that momentum is a vector and kinetic energy is a scalar, thus it proves nothing.
Specifically, what happens if we discuss scalarized momentum? What happens if we discuss vectorized...
This is from Taylor's classical mechanichs, 11.4, example of finding the Lagrangian of the double pendulum
Relevant figure attached below
Angle between the two velocities of second mass is
$$\phi_2-\phi_1$$
Potential energy
$$U_1=m_1gL_1$$
$$U_2=m_2g[L_1\cos(1-\phi_1)+L_2(1-\phi_2)]$$...