Work done Definition and 1000 Threads

Hello, everyone. I'm not sure if I did this problem right because there's no answer key. I got this question from a high school introductory Physics book. But it's jaded and I accidentally poured coffee on it, so I can't take a picture of the question. I copied it word for word though. This is...

Taking v = 2m/s, u=0m/s and s = 15m, we get ##a=0.13m/s^2## ##F_g = mg = 100(9.8) = 980N## Since there's no vertical acceleration, the normal force is equal to the weight ##N = 980N## ##f = \mu_k N = 0.05(980) = 49N## ##F_{net} = ma = 100(0.13) = 13N## ##F_{app} = F_{net}+f = 62N## My...

So this is a sketch I made of the situation and this is my approach my approach is incorrect , and Idon't seem to find the mistake , maybe B*p isn't correct. Any ideas?

My question is, is this correct, and if so, why the minus sign?

I first worked out the buoyant force using rho = 1000, g = 9.8 and V = 1. (this gives Fb = 9800) I then worked out Fg of the tank (5 x 9.8 = 49) Then I used Fnet = Fb - Fg - T = 0 to work out tension as 9751. I thought the work done would be the work done by tension (9751 x 50.4), but I'm not...

How would you go about calculating the work done in morphing a square current-carrying loop into a circular current-carrying loop, without change in length while maintaining the same angular orientation with an external magnetic field. My book suggests defining P(potential energy) = M.B (dot...

The question is given in 3 parts. For first part, process is isochoric so Work done=0. We know here that at end of the process (a), T2=T1 while V remains constant (we can take it as V1) so P2=2P1. For second part, process is isothermal so T is constant. At end of process we reach P1 again from...

Missed units in the photo - J Answer is 110 000 J

hi guys i have a conceptual question .As you know equipotential surfaces is one on which all point are the same potential there is no work required to move a charge from one point to the other . So my question is how can we change the locotion of a particle without using any force ?

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As stated, part (a) says that the work done by the gravitational force ##\vec{F_g}## is 59 kJ. If ##W_T## is the work done by the elevator cable during the 12 m fall, then using the work-kinetic energy theorem, \begin{align*} K_f -K_i &= W_g + W_T\\ \frac12m({v_f}^2 - {v_i}^2) &= 59000 + W_T\\...

I took a stab at the question, but I don't think I did it right. I know that Work = Change in Energy thus, Work = final energy - initial energy Because there is no energy at the final position, then final energy = 0 (I'm not sure if this is correct). So I got the equation W = 0 - (1/2kA^2 +...

W=-qEd =-(1.6*10^-19)(23)(0.75) = -2.76*10^-18 J However, the answer is 2.76*10^-18 J. Why is the word done positive and not negative? Since it's traveling in the same direction as the electric field, shouldn't it be negative work?

height to point A is 65m. no friction. comes to rest at point A. total mass of the roller coaster is 650 kg. the work-energy theorem states that the net work done on an object is equal to the change in kinetic energy of the object. kinetic energy is E=1/2mv^2. initial velocity is...

so the result I got is the energy lost by the field right??

The answers for author's solution differ slightly from mine when I convert his answers to SI. I think this is only a conversion issue as the differences are small but I wanted to confirm if my solution is correct: d=rt (distance = rate x time) so (12...

I know how to solve this using work energy theorem but how can one solve it starting from the fact that net force is centripetal? My attempt was "this means we only need an "anticentripetal" force of same magnitude, but centripetal force does 0 work hence new force also 0 work to counter it"...

Juri is tugging her wagon behind her on the way to... wherever her wagon needs to go. The wagon repair shop. She has a trek ahead of her--five kilometers--and she's pulling with a force of 200 Newtons. If she's pulling at an angle of 35 degrees to the horizontal, what work will be exerted on the...

Is my solution correct? $$ F(t) = at + b $$$$W=\int_t F(t)dr, ~~~~~~~~~ ~~dr = v(t)dt$$$$W=\int_t F(t)\cdot v(t) dt$$ $$ f = \frac{dp}{dt}$$ therefore $$v(t) = \frac{1}{m}(at^2/2+bt)$$then $$W = \int_t \frac{at+b}{m}\cdot\left(\frac{at^2}{2}+bt\right)dt$$ $$W = \frac{1}{m}\int_t...

first to find the force by gravity, it would be 4 kg * 9.8 m/s = 39.2 N then solve for work using W= F*d*cos(theta) W = 39.2 N * 3 m * cos 28 = 103.83 J My confusion is do I use sine or cosine and what angle do I use, the actual angle of the incline or the angle between the mass and the...

first I tried to use gravitational force to solve it but because of the incline the gravitational force is zero. so I used normal force to solve. Normal force is 39.36, then f*d*cos(theta) 39.36*3.6*cos(55) it would be cos 55 because the angle is in the direction of normal force from the...

I'm currently studying Method of Images in Griffiths book and in section 3.2 he introduces the method of images for a point charge at a distance ##d## from a grounded conducting plane at potential ##V = 0##. In subsection 3.2.3, Griffiths compute the energy of the real system and the image...

Work done if centre of mass is away from us will be less. But how? Or should we consider torque?

W= Fd W= 3NX1.20m W= 3.6 J is the work done by the crane W=Fd W=0.6Nx1.20m W=0.72 J is the work done by the car I have no idea how to solve the c, I am thinking at first that the work done by the crane and the car are zero because cos 90 is zero and there are perpendicular to each other. I...

Hi gentlemen I want to know the classification of thermodynamic processes according to which is higher than which in work done during either compression or expansion. Thermodynamic processes like isentropic and isobaric. Thanks

V = V(n,P) PV = nc dV = (-cn/P²)dP + (c/P)dn dW = (cn/P)dP - (c)dn This was how i tried to attack the problem, but it will end with log and probably it is wrong, any tips to fix it?

Homework Equations Conservation of energy. Kinetic energy=(1/2)mv^2 The Attempt at a solution m=1.5*10^3kg, final velocity=25m/s, initial velocity= 11m/s, distance=0.20 and work done=? What do I do next.

Here's a diagram of the object My Question: My friend and I were solving this problem. We both found Fxnet and Fynet. However to find work I did W= Fnet*d whereas he did W= Fxnet* d. He claims that since it's sliding on the floor it can only go in the x-direction, and it can't go in the...

Hi, A quick question on a conundrum I seem to have encountered. My main question is: why is it wrong to use the formula above instead of the SFEE approach? My approach: Use the formula: $$ w = \frac{R}{1-n} (T_2 - T_1) $$ From the data book, ## R = 0.287 ## kJ/kg K and ## n = \gamma = 1.4 ##...

Most companies specify full-time jobs as working 8 hours per day, or 40 hours per week. Some of these jobs have strict time schedule like between 9-5, others have a more flexible time. My question is more about the jobs with flexible time, since these jobs allow you to stay in the office for as...

we know ##W_g = -\Delta U## but here to find ##\Delta U## we will need another equation won't it be wrong to write $$-\Delta U = -\int_1^{0.8}mgdy$$ as this equation is derived from ##W_g = -\Delta U## and as we have 2 unknowns we will need two equations. this is a rather easy problem but I am...

I am not able to follow the derivation of work done in a reversible and irreversible process as I don't get why the work done should be different in the two processes. a reversible process is said to be a process that occurs infinitesimally slowly and an irreversible process goes from initial to...

The first law of thermodynamics states that the change in the internal energy of a system equals the net heat transfer into the system minus the net work done by the system. In equation form, the first law of thermodynamics is ΔU = Q − W. in the mathematical expression of the first law of...

Push-ups take a certain amount of work. If starting from a 45 degree angle instead of from a zero degree angle (the ground), my calculation shows it requires 1/√2 the amount of work. If the relative movement of the body, remaining stiff, is the same, the center of mass moves a certain height...

I do not understand how to get ##W = 32r^2##

I am a bit confused on the definition/convention of work. In some books I see statements that say : "If work is done on the system, its sign is positive. If work is done by the system, its sign is negative." And in other books I see things like: "By convention, work is regarded as positive...

There’s a current i in the loop in the figure. The Ampere’s force iLB on a wire of length L exerts on charges in the wire but it does no work on the charges. The charges would go in circular motion if there were no wire. Then the wire exerts exactly iLB on those charges to keep the charges...

I have drawn a picture of what the induced electric field will look like, and I have determined its magnitude both within and outside of the magnetic field. I was able to get the right answer for part (b) with this information, but I don't understand why the answer for part (c) is 0 J. It...

I have already calculated full charge inside the sphere: e = ∫ρ dV = 2πBr^2 And I know that electric potential on the edge of the sphere is: U = e/ 4πεr The idea is that I calculate work by the change of electric potential energy, but to do that, I have to calculate electric potential energy in...

This is the diagram in mind Some of my assumptions like the particle q will be moving with some initial velocity V1 in the circle of radius of r1 and when it moves to the circle of radius r2 its velocity will be V2 ( > V1) since its potential energy is decreased. The amount of work done by...

It was just a conceptional question. I thought that the positive work is done while it arises and negative work is done while it falls. For the effect of air resistance, the work done on each case would be less than the work done in absence of air resistance. Is there any other things that I can...

To do this I'm given a graph showing gravity reducing non-linearly from 3.7 N/kg at the surface to 0.7 N/kg at 8 x 10^6m. I believe that the equation for a fixed gravitaional pull is E = M x G x H - however with changing gravity as the object rises - how do you calculate this?

The diagram for the problem is shown alongside. In the vertical (##\hat z##) direction we have ##T \cos \theta = mg##. In the plane of the pendulum, if we take the pendulum bob at the left extreme end as shown in the diagram, we have ##T \sin \theta = \frac{mv^2}{r}## (the ##\hat x## axis of...

Hello all. Right now I am taking physics 1 and were doing the Work-Kinetic Energy Chapter. I was just reading the derivation for the work done by a spring's force. I understand how to get the result but what i don't understand what to make of it. I understand that because a spring's force...

I have been reading older posts containing "work done", but they don't seem to answer my doubts, or maybe I don't understand the answers because I need some visual pedagogical aids. I can sort of take the concept in, when we are dealing with a charge in an electric field, but when it comes to...

If the first equation for force is not formatted please refer to the pic. I did not post my full attempt as the problematic equation leads to a complex integral. Should I be converting r vector into its univector. The two outlined equations refer to the force. I am making an error transforming...

Why work done is area enclosed by graph of F v/s x on x Axis but not y axis. Suppose we apply a force on object which is proportional to displacement as ##\vec f##=## \vec x##²then area enclosed by Force and displacement on x Axis is integral of ##\vec x##²but on y-axis it should be integral of...

Here is the question: I have correctly calculated the power produced by the thrust force (P = Force x Velocity = 9.0 × 104W) , the work done by the thrust force over 3 minutes (W = Power x Time = 9.0 × 104 × 3.0 × 60 = 1.6 × 107 J) , and the gain in potential energy over this period (mgh =...

When we say that we apply force to a wall then we didn't transferred energy because of no displacement, but I think we somehow increase many types of energy of the wall. For e.g:- potential energy of particles in wall e.t.c. So according to me it means that we defined work only for mechanical...

Hi there I have been attempting the parts to this question and I'm finding some trouble on how to answer the last part which is d)iii Here is what I have done for the rest of the parts and what I think I should start off with in part d)iii Thanks!