Work Definition and 998 Threads

Let's say we have a cylinder of volume V1 with a frictionless movable piston and some gas trapped inside with pressure P1 and temperature T1. On top of the piston lay some small pebbles that add weight and essentially create the pressure P1. Also the system is inside a reservoir of water that...

I think I have an idea of this but I just want to be sure, and I also have some extra questions that I want to clear up. The work done on an object is the dot product of the force vector with the displacement vector, Fd*cos(theta). Am I correct in that this is without regard to velocity? i.e...

I have this little doubt because in class, the professor said the work is equal to the integrate of vdP, but i don't know how to prove it, why it isn't PdV like those exercises in a cylinder-piston

Consider this situation. A mass $$m$$ is dropped onto the spring system from rest , and it is given that the smaller spring gets compressed by a distance $$h$$ My main query in this question is regarding the use of Work Energy theorem, which states that the work done on a body by all the forces...

At first I tried using U1 = 1/2kx^2 from -x to zero then U2 from zero to +x but I don't know shall I add U1 and U2 or subtract them?

I wanted to make sure these were the correct equations to use for they were not given. I had to research to get this far and only had the below statement to go off. The problem is an object with initial velocity = "0." It travels 0.495m in 0.1625sec and weights 8.4 kg or Initial velocity (u) =...

the answer just takes wd = 1N x 20m = 20J. My question is why does it not account for the weight? In physics, when we say work done by the upward force, do we just take that force or must we calculate what the net force is? the next part of the qn then asks, "gpe + ke = work done by this...

Okay, so the first law ##dU=dQ+dW##. We all know that ##dU=C_VdT##. So that means that we have: $$dQ=C_VdT+PdV$$ Now I have a problem. We also have that $$dU=\left(\frac{\partial U}{\partial V}\right)_TdV+\left(\frac{\partial U}{\partial T}\right)_VdT$$ and substituting that in to ##dQ=dU+dW##...

By applying some simplifications, the power ##P## of a body moving at a speed of ##v## at an incline of ##\alpha > 0^\circ## can be expressed as: ##P = P_{acceleration} + P_{friction} + P_{gravity} + P_{air} = (ma + mgsin(\alpha) + C_{rr}mgcos(\alpha) + 0.5C_dA\rho v^2)\cdot v## If we would...

a) 0 b) (P2-P1)V c) Cp(T2-T1) d) Cv(T2-T1) < Ans I don't believe its B because if volume is constant, there's no work. I mostly don't understand why Cv is chosen instead of Cp.

This is chemistry but it's basically physics :D. I used PV = nRT, I get V = 37.44 L. This is fine. So then I have W = P(Vfinal - Vinitial). Vinitial is zero, because there was no hydrogen gas initially. So I get 3.78 kJ. And as the gas expanded from 0 L to 37.44 L, the gas has done positive...

Can someone help me by taking a look at the attachment and figuring out where I am making a bad assumption? It's driving me nuts.

We know that net work done is equal to the change in kinetic energy, so we write: $$\Delta K = W$$ The tension is acting at angle ##\theta## due to the x axis, so we will only be taking its x component ##T_x = T \cos{\theta}##. Since we can look at this as one dimensional motion (##T_y## does no...

I did classical mechanics a while ago and I was going over some stuff that I wasn't sure if I understood correctly and now I've come over this one. It says that work done by conservative forces is equal to the negative difference in potential energy. Or, ##W_c = - \Delta U##. And I've really...

I have found the answers for T = 848.615K, P = 126137.7705 Pa and change in S = -184.27008 J/K. But my answer for work is not correct and I am not sure where I am making an error. Could someone please help me out with how to calculate work? My steps for work is : We are given: Pi = 101325 Pa Pf...

Reference textbook “The Physics of Waves” in MIT website: https://ocw.mit.edu/courses/8-03sc-...es-fall-2016/resources/mit8_03scf16_textbook/ Chapter 2 - Section 2.3.1 [Page 45] (see attached file) Question: In the content, it states that we need to use real force and real displacement...

My answer is (D) but the answer key is (C). My reasoning is that the net work for (A), (B) and (C) are all zero since the object moves with constant speed so the resultant force is also zero. What is the mistake in my reasoning? Thanks

Ok, so currently, I'm working on problems involving work done by a general variable force. I had a question as far as solving these that ties into other problems as well... I see in some of the worked problems in this textbook, when they take the integral of a force equation to determine the...

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...

my attempt: centre of mass of standing block is at 0.6 and the fallen block is at 0.25 so then change in height = 0.35 and mg = 4000 N so change in gpe = 4000* 0.35 = 1400J but wouldnt we take into account that we only have to apply force till the centre of mass is outside of the block so...

my answer was c but i didn't understand why the work done in 16m or 12m wont be the greatest? how would we calculate work done in 12 m and 16 m??

my attempt: i solved it all correct but i dont understand a few things mentioned above... 82.04 * v = 56 so i got v as 0.68 m/s which is correct but i dont understand the concept...

I did: Work done by gravity+work done by applied force= KE(final)- KE(initial) Work done by gravity should simply be -mgh =100*5=-500J For work done by applied force we know: W=∫F⋅ds which can also be written as W=∫Fdscos(θ) since F is constant i can take that out W=Fcos(θ)∫ds here since ive...

I was just wondering if I needed to set up this double integral or not. I've never seen anywhere people setting up double integrals for calculating the work.

So for Q1, I answered down (towards Earth) but the solution says there is no acceleration there. For Q2, I answered mgh, but the solution says it's mgh/t, which is power, right? I just want to make sure I'm not super confused. Thank you.

these notes on magnetic field from MIT OCW's 8.02 course. There is the following snippet on page 10 At this point it is still not clear to me exactly what a magnetic dipole is. I've done all the calculations for a current-carrying rectangular loop, so I will assume that a magnetic dipole is...

A diagram is also provided, which looks like this: I'm not sure what is correct and after doing some digging online I still haven't been able to come to a consensus. I'm currently stuck between one of two possibilities: positive or zero. I'm pretty sure work on A (Wa) and work on B (Wb) are...

Im confused I did 850*9.8*14.7 = 1.2x10^5 Why is it not negative since the force is opposite to the motion?

For A the 1.2 kg block is being pulled by gravity hence work is done downwards which will make work positive since it's going with the same direction as the force. 1.2 * 9.8 = 11.76 N pulled downwards Work = F*d 11.76*0.75 = 8.82 J The tension is the other force and since the thing is...

Thus $$\Delta K=-\Delta U + W_{nc}$$ $$W_{nc}=\Delta K+\Delta U=\Delta E_m$$ My question is about the following statement The system is closed. ##\Delta E_{system}## does not necessarily have to be zero. Where does (1) come from?

The definition of work and power done over a continuous body is: $$ W = \int Tn \cdot u dA + \int b \cdot u dV $$ $$ P = \int Tn \cdot v dA + \int b \cdot v dV $$ ##T## is the stress tensor, ##b## is the body force, ##u## is the displacement vector, ##v## is the velocity, ##n## is the normal...

The answer key claims the answer is -0.9 J. By my math it should be -0.501 J, is that incorrect?

My first problem is to find the absored and rejected heat. Can I say that it is equal to the work done in an isothermal proccess (##dQ=Pdv##)? My reasoning : We have ##dQ=C_V d\theta + Pdv##. For constant temperature it becomes :$$dQ=Pdv$$

Question: Is it meaningful to think of the repulsion of mutual color charge and the attraction of three different color charge in QCD as being indicative of the classical concept of work taking place? Exactly, how is this explained in the context of three charges needed to elicit the...

To boil down the question, if you have a body at rest and apply a constant force, it will accelerate and the work done on it will be F*s (or the integral version of that statement). However, as the body accelerates due to the force, does that mean, per a given time unit, more and more work will...

i'm copying from the book... Hookes Law - F = -kx W = Fdcos∅ since ∅ is 180°, W = -Fd = -Fx W = ∫(-Fxdx) now the book says, from Hookes Law equation "the force magnitude F is kx. Thus, substitution leads to W = ∫(-kxdx)" why are they saying to substitute the magnitude of the force and not the...

I was trying to solve a JEE ADVANCED 2020 paper 1 question 13 on thermodynamics. I attempted the problem and got a different answer than they did. I did it in a different way, but I can't find a mistake in both methods. So now I am stuck. I think the problem is in the first step probably using...

Does the work "extracted" from a gas (with the same initial properties) against a piston while expanding change based on the mass of the piston? For example, I have a specific volume of compressed air inside a cylinder with a piston positioned horizontally with stops. The air temperature is...

This was the question This is my solution The problem arose after reading this post on PhysicsSE and this answer given So If I remember correct work done is ##-P_{ext}\Delta V## [FONT=times new roman]I don't understand why $$\Delta H=\Delta U+(5×4-1×10)L.atm$$ If that answer (the answer on...

I started by calculating the energy at the Earths surface, which is just -G(150)(6 x 10^24)/(6400 x 10^3) = -0.9 x 10^10, and calculating the energy at the orbit radius, E = -G(150)(6 x 10^24)/(18000 x 10^3) = -3.4 x 10^9, then doing some subtraction we have -3.4 x 10^9 - (-0.9 x 10^10) = 5600 x...

Because one is effectively pushing in the same way as a chest press machine, wouldn't the work done be the same?

HI, I need help wtih my physics work and dont know where to ask

I am posting this question after I thought I had easily solved the problem, but then when I checked the back of the book I saw that I was incorrect. Here is what I did. (a) $$W=-\int_{V_i}^{V_f} PdV\tag{1}$$ $$V=V(P,T)-\frac{nRT}{P}\tag{2}$$ $$dV=\left ( \frac{\partial V}{\partial P}\right...

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...

Is the following quote accurate: "The fundamental property in thermodynamics is work: work is done to achieve motion against an opposing force" Specifically, I am asking about the portion after the colon. I am a little confused by the notion of an opposing force. Let's say we are in outer...

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...

I tried E =Fxcos0 but only ended up with 243kJ

I have to show that the force is non-conservative, i.e. that work done for the round trip ##\neq 0##. Rearranging the circle equation, I can say that:$$x = \sqrt {R^2 - y^2}$$$$y = \sqrt {R^2 - x^2}$$ Then I have:$$F_x = -\sqrt {R^2 - x^2}$$$$F_y = \sqrt {R^2 - y^2}$$ Now, as I understand it, I...