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.
In textbooks or in lectures it is said that : Work can be either positive or negative: if the force has a component in the same direction as the displacement of the object, the force is doing positive work. If the force has a component in the direction opposite to the displacement, the force...
I was reading the chapter kinetic energy and work where I saw a derivation in which it was proved that work done=change in kinetic energy, so i understood from this derivation that work done is energy transfer by means of force but in this book kinetic energy is introduced earlier so as to find...
I'm asking for clarification, but it's my understanding, that of the thread below, and my college physics book Paul A tipler, that when walking or driving a car, the force of friction from the ground does no work. This makes sense in a car becuase the engine/fuel makes the power, driving the...
I found two formulas to calculate the work done. One is with this path integral:
## W_{AB}## = W(## r_A,r_B ##)=q* ## \int_{r_A}^{r_B} E*dr ##
but here is the one I tried to use:
## W_{AB}## = q*Δ U = q*(## \frac {kQ} {r_A} ## - ## \frac {kQ} {r_B} ## )
Now here's my problem, what are...
From what I understand, constraint forces do no work because they are perpendicular to the allowed virtual displacements of the system. However, if you consider an unbalanced Atwood machine, in which both masses are accelerating in opposite directions, you'll find that the tension force of the...
Apologies for my fairly undescriptive title, I'm having trouble coming up with a better one.
I'm simply trying to verify whether my understanding of this problem is correct.
1. Pushing the oven up a ramp 4m long would require more force (my reasoning here is based on the results of a lab which...
I'm having a little trouble with part a) of this question:
Since it is stated that the heating is slow, I thought it was reasonable to assume the process is reversible which means that the pressure in both sides should be equal. Consequently, $$W = - \int_{V_{0}}^{V_{1}} P dV = -...
The proof of magnetic forces do no work is given in Introduction to Electrodynamics by David J. Griffiths like this
My problem is why he has replaced d\mathbf{l} with \mathbf{v}dt? This substitution implies that the charged particle was moving with \mathbf{v} only and no force acted on it...
Since the assignment asks the work done by the gas, that should be equal to P1*(V2-V1) aka the area under the P1 line. Do I have to subtract the work done to the system or is this the solution already? If so, why do I need P2?
Hi,
Could I please ask where I am going wrong with this very simple question:
Here's my answer (units implied):
A force of 20 extends the spring by 1/100 and so the Work Done in performing this extension is 20 * 1/100 = 1/ 5
Now, the work done in extending a spring is given by the formula...
A box is moving at a constant velocity. 250N of applied force is acting on a box while 250N of friction is acting on it on the opposite direction. If the box travels 4 meters in the same direction as the applied force, how much work is done?
I know that work = force * displacement.
Is the...
if I shoot the football, the ball will recive a force in specific direction, and it would follow that direction.
Does the work is done by me when I shooted the ball, or the ball does work when it moves all the period after my shooting ?
the work includes force and displacement , so who does the...
My teacher told me that answers are F=1200 N for lifting straight up and F= 360 N for using the ramp. I can get the force for lifting straight up by using 1800=F(1.5)cos(0) but I do not understand why for using the ramp the equation 1800 = F(1.5)cos(72.5) does not produce the correct answer.
So I've been digging this problem for quite some time, and still can't figure out a way to obtain work done in AB.
I do understand that work done in AB = area under the graph. However, I can't figure out how to obtain the volume at B. I can't use PV=nRT since I am unable to obtain the...
Attached is an example from my Tipler physics book of a determining the power delivered from a cars engine. I get the derivation(math), but am having some issues with the understanding/concept.
Mainly, isn't Power=Work done/time. If so, they define the system as Car-earth. There is no external...
Fa = Force applied on the cart by the parent
Fax = x component For the force applied on the cart by the parent
Fay = y component For the force applied on the cart by the parent
Ff = force of friction
The method I thought to use was to solve for Facos, Fasin and then use pythagorean's theorem to...
https://www.physicsforums.com/attachments/250022
For the part (a), set the tension of the string that pulls the elevator be ##T_1## and the tension that pushes the counterweight be ##T_2##. Then we have the following equations:
$$T_1−1200g=1200 \cdot 1.5$$
$$1000g−T_2=1000 \cdot (−1.5)$$
where...
Applying the dot product to the position vector and force vectors yields the result that the force vector is always at 90 degrees to the position vector, namely that it is directed tangentially.
The mark scheme gives the work done as 0, however I don't agree. I believe they got this by finding...
I have a Hamiltonian ##H_{\lambda(t)}##, where ##\lambda (t)## characterizes a time dependent path in parameter space. The parameter is changed in finite time from ##\lambda (t_i)## to ##\lambda(t_f)## . At ##t = t_i## the system is in the intial state ##|\Psi>##. What is the work done on the...
I am reading Landau's Vol 5 on Statistical Physics and have trouble grasping some concepts in Section 20.
If I understand this correctly, the body and the medium are in direct contact and can exchange work and heat while the object can only exchange work with the body.
So the minimum...
I was reviewing physics stuff and I've managed to confuse myself.
I was reading about work (principles of ohysics, serway&jewtt):
-work is an energy transfer
-if work is done on a system and W is positive, energy is transferred to the system; if W us negative, energy is transferred from the...
My book states that, '∑Finternal=0∑Finternal=0. But the work done by them may or may not be zero.
Why is this true?
Edit: How can the work ≠ 0 sometimes?
Okay, say we have two balls(equal mass and size), 1 and 2. #1 has kinetic energy and #2 is at a standstill, they collide. Ignore all friction, heat, sound losses etc..
Now I know that 1 exerts a force on 2, hence doing work, which in turn uses its kinetic energy up. Therefore, 2 speeds up...
So I have honestly no idea what kind of force is acting on the objects in each case, nor if work is being done. I attempted to identify the force for ‘d’ and I wrote the following: “work is being done and forces acting on the book are balanced, the weight of the book acts downwards and the...
Work done in moving a charge around the circular path is 0 because the distance between the charges remain the same so there is no potential difference and delta(V)=0 but I thinkk some torque or external force must be required to move the charge around the circula path
I calculated friction = 1N.But since force is more than maximum friction,therefore lower block will not move and a force of 0.5N will be applied to upper block.
Therefore its work done by friction will be W=0.5×6=3 J.
But answer given is 1 J.
Can anyone help.
Suppose a metal disk is spinning at constant rate of one cycle per second and at a radius of about four inches a small metal block is pressed against the disk. The circumference of the path of the contact is exactly one foot.
The coefficient of kinetic friction between the disk and the small...
Dear Physics Forums people,
My problem lies in understanding how the following line integral, which represents work done by the gravitational force, was calculated
Specifically, in the integral after the 2nd = sign, they implicitly used \hat{r}\cdot d\vec{s} = dr
I wish to understand what...
$$W = - \int _ { a } ^ { b } \vec { F } \cdot d \vec { r }$$
( The Force here is referring to the applied force, When moving a positive charge towards another positive charge(stationary) / field
In this formula why there is a negative sign in the formula? I am not asking the sign of the total...
I have a quick question about the work done concept here, especially the line integral part of it. So I understand the fact that the work done from getting from point A to B is: \int_{a}^{b} \vec F \cdot d\vec r .
However, within the context of electric fields, when we define electrostatic...
we know that if the applied force is in the direction of the displacement then work done is positive.But in case of bringing 2 opposite charges from infinite to a certain distance,the work done is negative even the force and the displacement of the charge is in the same direction.
From...
I already solved the first part, but according the my book, the answer isn't quite correct. This is what I did.
Finally, I ended up with ##a=\frac{F(r-R\cos\alpha)}{Rm(\gamma+1)}##. But according to my book, the answer is ##a=\frac{F(\cos\alpha-\frac{r}{R})}{m(1+\gamma)}##, what am I doing...
I was just doing some review on my physics lecture and I stumble on the idea of what if there was an object hanging and the cord mass is also included in the weight and it's displaced upward without having velocity nor time hypothetically and the cord change mass. I tried solving it by W = fΔx...
Homework Statement
A cart of mass M1 = 6 kg is attached to a block of mass M2 = 3 kg by a string that passes over a frictionless pulley. The system is initially at rest and the table is frictionless. After the block has fallen a distance h = 1 m:
What is the work Ws done on the cart by the...
Homework Statement
Two blocks are attached to a rope across a pulley. Block A is on the ground and has a mass of 4 kg. Block B is 2 meters above the ground and has a mass of 12 kg. When the blocks are released, block B hits the floor and block A rises to 2 meters; each block has a final speed...
An open tank has the shape of a right circular cone. The tank is 8 feet across the top and 6 feet high. How much work is done in emptying the tank by pumping the water over the top edge? (The weight-density of water is 62.4 pounds per cubic foot.)
Homework Statement
A sledge loaded with bricks has a total mass of 18.0 kg and is pulled at constant speed by a rope inclined at 20.0° above the horizontal. The sledge moves a distance of 20.0 m on a horizontal surface. The coefficient of kinetic friction between the sledge and surface is...
Homework Statement
a) A point charge + q is placed at the origin. By explicitly calculating the relevant line integral, determine how much external work must be done to bring another point charge + q from infinity to the point r2= aŷ ? Consider the difference between external work and work...
Hi all, i am trying to figure out if work done by gravity is positive or negative on an approaching meteor,
I have used W(1->2) = integral (1->2) F(gravity) dot dr. and got a result of (GMem/r2) - (GMem/r1).
Were G is gravity, Me is Earth mass, m is meteor mass and r1, r2 are initial and...
I am stuck on like the first page of thermal physics. It seems like the signs of the work done on the system are opposite when the volume is expanded and compressed. But when I imagine myself pushing or pulling the piston, I get confused from W = \textbf{F}\cdot \textbf{d}
This work will be...
I'm trying to determine the work done by a person as they pull a luggage up a ramp. The ramp has a height of 5 m and the distance the person walks up is 20 m. The weight of the bag is also 10 kg.
I am trying to compare the work done by pulling the luggage up a ramp to carrying an equally heavy...
Work W done by moving the object with force F for distance s is W = Fs.
When I move the same object the same distance but with twice the acceleration, does
the work done gets also doubled?
By F=ma, doubling the acceleration yields m*2*a = 2F -> 2Fs = 2W.
I've mostly read, that if I want to...
Homework Statement
https://imgur.com/vQRoMjZ
Homework EquationsThe Attempt at a Solution
I believe that the inductor does negative work on the charges via the electric field (by way of induced EMF).
I am unsure about how the capacitor would be doing positive work, though. Can anyone help steer...
So the work done when charging up a capacitor is ##dW=VdQ##
However, when we add a charge ##dQ## to the capacitor, ##V## also changes accordingly, so I was wondering why the work done wasn't written as ##dW=VdQ+QdV## (one that also takes into account t he change in ##V##).
Thanks in advance.
Are the following definitions correct?
Work done as energy transferred
The energy transferred when the forces between two objects interact
Work done by a force
Work done = force x distance moved in the direction of force applied (W=F*Δs)
Work done by a gas
Work done = pressure x change...
I' m trying to derive the work done by a torque from W = ∫ F ⋅ ds and I' ve looked up the internet, it said:
W = ∫ F ⋅ ds ( since ds = dθ × r ) ---------------------------------------- ( Line 1 )
it can be written as
W = ∫ F ⋅ dθ x r
this is a vector triple product , thus can also...