In physics, work is the energy transferred to or from an object via the application of force along a displacement. In its simplest form, it is often represented as the product of force and displacement. A force is said to do positive work if (when applied) it has a component in the direction of the displacement of the point of application. A force does negative work if it has a component opposite to the direction of the displacement at the point of application of the force.
For example, when a ball is held above the ground and then dropped, the work done by the gravitational force on the ball as it falls is equal to the weight of the ball (a force) multiplied by the distance to the ground (a displacement). When the force F is constant and the angle between the force and the displacement s is θ, then the work done is given by:
W
=
F
s
cos
θ
{\displaystyle W=Fs\cos {\theta }}
Work is a scalar quantity, so it has only magnitude and no direction. Work transfers energy from one place to another, or one form to another. The SI unit of work is the joule (J), the same unit as for energy.
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...
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 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...
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...
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##
I don't understand why $$\Delta H=\Delta U+(5×4-1×10)L.atm$$
If that answer (the answer on the PSE post) is...
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...
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)
$$W=-\int_{V_i}^{V_f} PdV\tag{1}$$
$$V=V(P,T)-\frac{nRT}{P}\tag{2}$$
$$dV=\left ( \frac{\partial V}{\partial P}\right...
EDIT: I've finally found the solution, so here's what I did.
First, calculate Work using the equation: (F-mgsin(theta))*displacement, where F=the force being applied by the push, theta is the angle of the ramp, and displacement is the length of the ramp.
Now that you have the value for work...
Is the following quote accurate:
"The fundamental property in thermodynamics is work: work is done to achieve motion against an opposing force"
Specifically, I am asking about the portion after the colon. I am a little confused by the notion of an opposing force.
Let's say we are in outer...
Hello guys,
I need help on this problem,
"You throw a ball with a mass of 0.4kg against a brick wall. It hits the wall moving horizontally to the left at 30 m/s and rebounds horizontally to the right at 20m/s. (a) Find the Impulse of the net force on the ball during its collision with the wall...
I have to show that the force is non-conservative, i.e. that work done for the round trip ##\neq 0##.
Rearranging the circle equation, I can say that:$$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...
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...
This question was, effectively, asked here (please refer to that question for additional context); however, I don't think the given answer is correct (or at least complete) despite my having added a bounty and having had a productive discussion with the answerer there. In particular, I don't...
I didn't have much trouble with part a but I'm struggling with b,c, and d. I considered the efficiency formula for a heat engine e = work done by engine/ qh
but i am unsure of how to approach it.
for part c) not sure how i can get to Tc without knowing Th
for d) my gut is telling me 5/2 but i...
Hi everyone
I'd like to detect collisions between a circle and a floor.
I have set them up as follows:
The script runs fine (it's just a circle falling towards the floor), but nothing happens when the two objects collide.
Does anyone know what I've done wrong?Thanks
The question was this:
My calculations show that the answer should be equal to work done on crate to make it reach the same velocity which is equal to 216 J but the answer given is 432 J
It is believed that extra energy is needed to overcome friction but friction is an internal force and...
My question is whether I've formed the integral for the work done correctly? It just seems a bit unwieldy to me...
If I call the extension of the spring ## x ##, I can see that ## z = \frac l 2 + x ## and ## z^2 = \left( \frac {l} {2} \right)^2 + y^2 ##. Combining them gives: $$ x = \sqrt {y^2...