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.
Imagine a projectile with density p, initial velocity v0 and initial stagnation pressure = 0.5p{v0}^2 being fired into the gas chamber with a final gas pressure given by Pg = nkT, for n being the final number density before (at max pressure of gas) and T the corresponding temperature of the gas...
As a child, one of my family friends was an 80 year old neighbor, Harry R. Caldwell, who had been a famous tiger hunter in his youth. He killed tigers and Kodiak bears with a .22 or .25 caliber high power load that some felt too small for big game.
[edit: the claim about the bear is false...
I approached the problem using the work done by force equation (W=F*d)
In my understanding all potential energy would have been converted into kinetic energy (KE) by point P (no friciton)
We know d= L ; W= Wf ; f = Ff height = (H+h)
So the Energy at point P is entirely kinetic:
Which translates...
The total kinetic energy of a double pendulum can be calculated through the formule reported in the following article: https://dassencio.org/33
This works if the double pendulum in question is formed by two masses connected to each other and — one of them — to the point of origin by a...
Mentor note: Moved from a technical section, so is missing the homework template.
A Body of mass 1kg moves with a uniform velocity of magnitude 12m/s, a resistance of force of magnitude 6x^2 (newton) where x is the displacement (meter) which the body travels under the action of the resistance...
Suppose I accelerate a mass from rest to 1 m/s using n J of energy. From the sun's perspective, I've just accelerated it from (say) 29,785 m/s to 29,786 m/s, which would require 59571n J of energy. Where is the extra 59570n J coming from?
If the answer is "nowhere, changes in KE are relative to...
Hi all.
I'm trying to get my head back onto mathematical thinking and I'm setting myself projects to undertake to facilitate that. It's been too long since I've worked with numbers like this. My first task was to work out how much energy would be released when dropping things onto a neutron...
For part (c) my understanding is that in this case for the particle to cross the origin it must have at least as much kinetic energy as there is potential energy at ##U(0) = A##. Given the potential energy and kinetic energy at any position x=a is equal to ## U(a) = \frac {ka^2} {2} +...
This question is from the Jee Advanced 2017. This question stumped me because it was very difficult to do, the first part of the comprehension was asking for the kinetic energy of the ring.
This is what I mean.
Using the no-slip condition we can write ##\omega R-v_c=\omega_0r##. This is the...
TL;DR Summary: I cannot find any information on how to calculate min. KErot at which the gyroscope does not fall over
I am doing a project for school in which I investigate energy loss in a gyroscope. I apply a torque on a gyroscope to initiate its rotation and then measure the time it takes...
Hi,
I assumed I was supposed to find the amount of kinetic energy body 2 receives after contact, assuming the collision is central, body 1 will be at rest after the collision.
I started by using the equation for conservation of momentum:
\begin{align}
m_1v_1 = m_1v_1' + m_2v_2' \\
50 * 20 =...
Hello everyone,
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?
As the piston of an internal combustion engine moves up and down inside the cylinder, it goes from zero-max-zero speed. In a race engine, the maximum speed may reach up to 100 mph. So the question is how much energy is expended every time the piston decelerates to zero speed when it hits bottom...
The system of two material points of identical mass connected by a rigid rod of negligible mass and length ##L## is an example of a conservation of energy problem. The initial energy of the system is the sum of the kinetic energy of the two points and the potential energy of the rod, which is...
So i started off breaking up the problem into two sequences, right before the collision and after the collision has happened. I need to find the first ball's speed immediately before the collision which is no problem. PEi = KEf > mghi = (1/2)mvf (vf being the velocity right before the collision...
As far as I know, something like relativistic mass is just a concept, just a trick, there is nothing like the relativistic mass. When I move faster, I have higher kinetic energy, but my mass is still the same as if I was at rest. Kinetic and potential energies do not increase object's mass.
A...
I've tried to solve this exercise but I haven't used one of the properties of the system (the displacement of the masses) so I don't know if I'm wrong about my procedure.
First of all, we (obviously) know that
$$
P=P
$$
And since we can express the power of a force in two different ways, we...
For this question part d, KE=mv^2/2=q^2B^2r^2/2m (I rearranged B=mv/qr for v and subbed into mv^2/2). q^2b^2r^2/2m=2F_cyc^2r^2m(pi)^2
But when I subbed the values in I got 16.45MeV but the answer says 165keV instead. I'm not sure what went wrong?
What's a good explanation for part e also?
If we say that a proton has a kinetic energy of ##50## GeV, can we say that each of the three quarks that compose it have roughly a mean energy of ##\approx\frac{50}{3}=17## GeV?
If not, what can we say about the energy of each individual quark inside a baryon with a known energy?
Given that we're working with an elastic collision we want to populate the following system:
##k_{i} = k_{f}##
##p_{i} = p_{f}##
Solve for kinetic energy just before and after the collision:
##k_{i} = \frac{1}{2}mv_{i}^{2}##
##k_{f} = \frac{1}{2}mv_{f}^{2} + \frac{1}{2}I_{P}^{sys}...
EDIT: I've finally found the solution, so here's what I did.
First, calculate Work using the equation: (F-mgsin(theta))*displacement, where F=the force being applied by the push, theta is the angle of the ramp, and displacement is the length of the ramp.
Now that you have the value for work...
Hello! I am new to the differential version of classical physics, and I am trying to work how to derive kinetic energy from some pre-assumed equations:
Assume that we know: ##\ddot{z} = 0## and ##m\ddot{\textbf{r}} \cdot \dot{\textbf{r}} = 0##This results in $$\frac{1}{2}m\dot{r}^2 = W =...
Hello guys,
I need help on this problem,
"You throw a ball with a mass of 0.4kg against a brick wall. It hits the wall moving horizontally to the left at 30 m/s and rebounds horizontally to the right at 20m/s. (a) Find the Impulse of the net force on the ball during its collision with the wall...
We have ##U(x) = U(a) + \frac{1}{2}U''(a)(x-a)^2## (by taylor series)
It's known that ##U'(a) = 0## and point ##a## is a turning point, hence at that point, Kinetic energy is 0, and ##E = U(a)##, hence:
We have ##U(x) = E + \frac{1}{2}U''(a)(x-a)^2##I need to get equation of motion and I want...
note:
m = relativistic mass
##m_o## = rest mass
v = velocity of the objectQuestion 1: If a particle is moving at relativistic speeds what would it's kinetic energy be?
I think it's ##K.E. = \frac{1}{2} m_o v^2## and my friend thinks it's ##K.E. = \frac{1}{2} \frac{m_o...
I was working on this problem but after getting to the answer I questioned the methods that I used for previous problems that I had solved. I understand that the total energy of the system remains constant and that we use the conservation of momentum to relate the two velocities. This gives two...
I'm asking about this with particular reference to the signs of the answers. Here is the question:
The answers in the back of the book are:
(a) ##1.2 \times 10^4 \text{ N}##
(b) ##39 \text{ m}##
(c) ##4.7 \times 10^5 \text{ J}##
(d) ##4.7 \times 10^5 \text{ J}##
Here's a rough sketch of...
I tried to solve it using the work-energy theorem.The work done to make it stand on its one vertex should be equal to the change in its kinetic energy.
I am confused what will be the value of radius here? I have seen formula of kinetic energy for rolling of circular objects.Can anyone please...
i tried to calculate the the kinetic energy by using the impulse law and got that the speed is 150Ns/65kg then I calculated the Energy and later the power. My answer is was 433Watts. How would you solve this question?
Hi,
Am i correct in thinking that if we take a block of ice, moving at a constant velocity, it's then exposed to a heat source which melts the ice and turns it into water vapour, that we have simply removed any Kinetic energy, by Sublimation or converting it into heat.
My question is does the...
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...
Dear People,
I have a question. I have a rotating tube like a line that has two end and one of them is the center of rotation (like a watch arrow just tube), and inside the tube a mass that is moving towards the center of rotation. So the masses moving along the line aka along the length of the...
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}}$$
I've started off by calculating the...
I would guess that by multiplying the pressure exerted by the shockwave on the body, and then the resulting force - here ~69 Newtons - per the distance the shockwave passed through when traversing body A, I could get the work done but I’m not sure if it’s that easy and whether or not I should...
W_ext is the external work done on B and C, which is 12 J
Delta K_tot is the internal work, which is the work done by A on B plus the work done by A on C
Delta K_tot = 5
Solving for \Delta U, we find that the change in potential energy is 7 J
This answer says otherwise...
Dear Forum,
I am solving for the expectation value of the kinetic energy for the deuteron (Krane problem 4.3). I must be missing something since this has become far more complicated than I remember.
The problem is as follows:
## <T> = \frac{\hbar^{2}}{2m} \int_{0}^{\infty}...
If f(x) is the function of the "ground": My first assumption is that in a certain $$\bar{x}$$, $$f''(\bar{x})=0$$, and from that point I will analyse the situation.
The object has initial energy $$E_0=\frac{mv^2}{2}+mgf(x),$$ then
$$v=\sqrt{\frac{2}{m}}\sqrt{E_0-mgf(x)}.$$
In each point the...
First I found work:
W=(3.85x10^5)(2.45x10^8)
W= 9.43x10^13
Then used that for difference of kinetic energy:
9.43x10^13 = (1/2) (4.55x10^4)v2^2 - (1/2)(4.55x10^4)(1.22x10^4)^2
9.43x10^13 = (22750)v2^2 - 3.386x10^12
9.43x10^13 + 3.386x10^12 = (22750)v2^2
9.77x10^13 = 22750v2^2
9.77x10^13/22750...
I'm a little confused because my teacher used Bill's 500J of work for the kinetic energy equation and I don't understand why. I used the net work, so 300J, to find the speed and I'm not sure why that's wrong. Wouldn't friction make the wagon move slower than if there was no friction? So why...
hello guys, I wanted to ask whether I can just consider/think about this as being rotation around a fixed axis in a plane representing it as if it was 'just' a rod. This is mainly so that for the kinetic energy in the second position is where if we think about it in just a plane. Is this...
Kindly help me solve this question. The only thing so far that I know in this question is that energy is conserved and the momentum of Alpha particle will equal momentum of Thorium.
Spacetime expands at an accelerated rate and the particles with movement associated to this expansion are coupled to the Hubble flow. In many papers that I've read, objects coupled to the Hubble flow are treated as if they have some velocity and kinetic energy associated with it.However, can...
Since there is no friction : $$ 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?