# Tag Info

2

No the point of contact is not at rest. It moves with the block. You are probably confusing with rolling motion in which the point of contact is always at rest. There the point of contact is rest because the lowest point on the disk has two contributions, one due to forward motion of disk as a whole (v) and one in the backward direction due to rotation ...

1

According to Newtonian mechanics, it is true that the table exerts an equal but opposite force against gravity that results in $\Delta y = 0$, where $y$ is the up/down dimension. However, sliding block is clearly moving in the $x$ dimension (i.e. horizontally across the table). And it is also acted on by a force, namely friction. The block does not generate ...

0

The sled ($m = 11.1\;\mathrm{kg}$) shown in the figure leaves the starting point with a velocity of $25.1\;\mathrm{m/s}$. Use the work-energy theorem to calculate the sled’s speed at the end of the track or the maximum height it reaches if it stops before reaching the end. The straight sections of the track (A, B, D, and E) have a coefficient of friction ...

0

Try to keep this tidy. It is a straight forward calculation, but there are many terms, so tidiness is the key. Start with the Energy work relation: $$E_A - E_E = W$$ where $E_A$ is the energy at the beginning, $E_E$ the energy at the end and $W$ the energy loss due to friction. We have to split $W$ further into $$W = W_A + W_B + W_C + W_D + W_E$$ where ...

0

I am not so sure about your first step. One sees a gain in speed due to conversion of potential energy into kinetic energy. That would give ${{m}\over{2}} v^2_\mathrm{f}-{{m}\over{2}} v^2_\mathrm{i}= m g h$ no need for angular functions as the hight is already shown in sketch, but you must remove losses due to friction as well. Looking at the standard ...

0

To simply state an answer to each question: No displacement = no work done the definition of work is when a force acts on an object to displace it Yes, you can calculate the work done by each force. You can calculate the applied force pushing the mass one way, and the applied force pushing the mass the other way. Note that because there is no ...

0

In your first approach I believe you got a wrong sign. It should be $$v_f = \sqrt{v_0^2 - 2g d \sin(\theta)},$$ otherwise your speed will increase with distance. In your second approach seems like you are equating the work of the gravitational force to the difference of the mechanical energy. However, the work of a force is equal to the difference in the ...

-1

Work done is the scalar product of Force and displacement along the direction of the force, so work done should be zero since net force on the object is zero, hence the constant velocity. That means W = 0.

0

When the sloth starts falling, it's potential energy starts getting converted to kinetic energy, and just before it hits the ground, all of its potential energy has been converted to kinetic energy. Now, the problem you're facing is because of the fact that you're considering just the sloth, which is wrong, because the sloth is interacting with the ...

0

There is an external agent that removes mechanical energy from the sloth, namely the normal force exerted by the ground. You are right that no net work is done, but remember it is the work-kinetic-energy theorem: The net work equals the change in kinetic energy of the sloth.

3

Conservation of energy, as you note, holds for "the system." For instance, if you push on a ball, that ball gains energy, but the energy of the ball is not conserved--only the energy of you and the ball. In this case, the system needs to include more than just "the sloth" because the sloth is not an isolated system--there are external forces at work. Here, ...

2

Q = mc(t1-t2), Now, m = (density)(volume), Specific heat of water, c(in joule/gramCelsius) = 4.186, Hence, you can find the energy it would require for this conversion. . And the work you do can be a bit more pertaining to your efficiency.

2

You are sloppy with units, but the result is correct. To go from 25C to 3C is 22 cal/g. When you multiply by 300 g you have cal and your conversion to kJ is correct. Converting to W-hr is silly, but that is the unit of energy, not W/hr. You have 8.3 W-hr you want to remove. That chills the water assuming no new heat is added, so insulate the water. ...

3

One need not follow these steps. Indeed let $\gamma : I\subset \mathbb{R}\to \mathbb{R}^3$ be the trajectory of a particle. It's position at time $t$ is $\gamma(t)$, it's velocity is $\gamma'(t)$ and it's acceleration is $\gamma''(t)$. It's easy to see that $$(\gamma'\cdot \gamma')'(t) = 2\gamma'(t)\cdot \gamma''(t),$$ so the work done by the resultant ...

1

Formally, we have the standard dot product identity $$\frac{\mathrm{d}}{\mathrm{d}t}\mathbf{A}\cdot\mathbf{B}=\frac{\mathrm{d}\mathbf{A}}{\mathrm{d}t}\cdot \mathbf{B}+\mathbf{A}\cdot \frac{\mathrm{d}\mathbf{B}}{\mathrm{d}t}.$$ Inserting $\mathbf{A}=\mathbf{B}=\mathbf{v}$ gives $$\frac{\mathrm{d}\mathbf{v}^2}{\mathrm{d}t}=2\mathbf{v}\cdot ... 1 This might be better on the engineering SE site but here is some physics to consider: You are right the the heat you need to remove is the mass of water times the temperature difference times the specific heat capacity. Peltier devices and other heat pumps typically have a parameter called a COP - coefficient of performance. This compares their efficiency ... 0 Ok... I see the confusion... See when you bring a ball from infinity to a position at a distance r from the center of the earth with uniform velocity, you need to apply some external force on it ; otherwise naturally the ball would be accelerated (because of gravitational force of earth ) and the decreased potential energy would appear as the kinetic energy ... 1 I think what may be referred to in the question is the determination of gravitational potential energy (GPE). The definition for GPE is the work that was done in bringing the two masses to a distance r apart from an initial separation that was infinite. (Tsokos, 396) What is important here is that this happens at a very slow uniform speed so the Kinetic ... 1 When a ball of mass m is brought with uniform velocity from infinity into the g field of the earth at a distance r from it, the potential energy of the ball earth system decreases from 0 to -GMm/r. What does this lost energy appear as? Kinetic energy, which is typically radiated away into space. Imagine your ball starts off a long long way from Earth, and ... 0 Static friction does not produce or consume work in most of the times. For example for a solid body that rolls without sliding the velocity of the base point A is \vec v_a = \vec v_{cm} + \vec v_{tangential} \Rightarrow v_a = v_{cm} - \omega R = \omega R - \omega R =0 which implies that x_a = 0. The static friction is a force that acts on A so W_T = ... 0 It doesn't. The work is done by the active force(ex. A human trying to pull a bull.). This work is converted into frictional energy(ex. Heat generated b/w surfaces) 3 I suppose you read this passage in the famous Feynman Lectures. I am fairly certain that what Feynman is referring to (and what you are looking for) is a proof that an electrostatic field is conservative. There are a number of equivalent ways of stating that a vector field is conservative, each of which can be taken as a definition. Let \vec{F}(x) be a ... 2 In terms of the Young's modulus the Hook's law (up to the overall sign) is written as$$F=ES \frac{\Delta L}{L_0}=kx, $$the corresponding elastic energy (or the work that has be done to stretch a wire) is$$W=ES \frac{\Delta L^2}{2L_0}, $$here you know the Young's modulus E=1.3 \times 10^{10}, the cross area S=1.7cm^2 and the initial length L_0=0.89m ... 0 Without friction between the crate and the truck bed, the crate would remain at rest in the frame of reference of the road, as the truck accelerates away down the road. The crate moves in the frame of reference of the road, because of the force of friction acting on it. So the work done on the crate, in the frame of reference of the road, is the friction ... 2 I think you are confused about what d is supposed to mean in the equation W=F\cdot d. You seem to be under the impression that d is the distance that the object being acted on moves relative to the object providing the force. But this is not the correct meaning of d in the equation and you know it. Imagine if the car crate were in front of the ... 3 The contact point between the wheel and the car is stationary - there is no "rubbing" there (well there is because the contact point is really a patch but let's keep it simple). To do work you need "force times distance" - and without relative motion there is no "distance". When you apply the brakes in a car you have sliding of the pads relative to the ... 0 This appears to be a very normal intuition challenge: someone has claimed the math says the world works one way, and intuition declares it works the other. Let's see if we can get intuition to agree with the so called "math." Gravity is a sneaky thing. Even at the Quantum Mechanical level, it keeps causing problems. It will be easier to understand if we ... 0 Yes, the net force is zero. But that doesn't mean you are doing no work "against gravity". That the system is in stable equilibrium, does indicate that the kinetic energy is constant. You are changing the configuration of the system- its state of position - against a field where the force is given by the gradient of potential energy ie.$$ \mathbf{F} = ...

1

As you say, there is a downwards force on the plane of $Mg$ where $M$ is the mass of the plane and $g$ is the acceleration due to gravity. There is also a force due to aerodynamic drag, but let's ignore that for now. The key to understanding how the plane stays up is that force is equal to the rate of change of momentum. In this case the wings of the ...

1

It is zero; think of the conservation of energy. Actually, because of this law, all force's work on an object will be equal to only its change of kinetic energy (at least in the Newtonian physic), and in this case this is zero. We use the potential energies to ease up the calculations (so you don't have to calculate all times i.e. the Coulomb-force's work): ...

1

No that's not right. a work can be calculated for each force individually. you are mixing it with the equation between work and energy that says: $$W = \Delta K \quad \text{K is kinetic energy}$$ but this work is actually the work of the overall force on the object. the work done by earth gravity force and your force is zero. But each is doing some work ...

-1

Assuming the plane is essentially a hover-craft that is gliding through the air without any resistance, the work must be zero over any distance it travels. The force that the plane is exerting to counter the force of gravity (and stay up in the air) is purely in the radial direction, while it is (assumed to be) moving in a direction perpendicular to it. ...

1

The plane itself is exerting some force in order to overcome the effects of gravity. Consider that if you turn off the engine, the plane would go crashing down. The only reason you are able to observe a constant velocity is because the plane is exerting some power to work against gravity (and other phenomena such as air-resistance etc.) Now you may be ...

0

You can change the sign of your angle, or you can swap a and b, but you can't do both. When you swap a and b the angle gets increased by 180 degrees (which is the same as changing the sign of your angle), if you do this and change the sign of the angle (again) the two cancel each other out.

2

You are using the repulsive force as the force acting to move the charge from B to A(which is not actually moving the charge). We need an external force to move the charge from B to A, which will be taken into consideration(to calculate workdone).

2

You are right. In Newtonian physics, work depends on reference frame. Force does not. Let's start with a book sitting on a table. The table exerts a normal force on the book, but it does no work because there's no motion. Next, imagine that the table is in an elevator and the elevator is going up at constant speed. The force between the book and table is ...

0

Just to add to Jerry Schirmer's answer, I find it helpful to think of a potential as being a more compact form of expressing $\vec F$ because $\vec F$ is 3-dimensional and $\phi$ is one-dimensional. If $\vec F$ is conservative, it would seem to only contain "one dimension" worth of information; it has some redundancy in its form (i.e. is not its simplest ...

2

the heart of a force being conservative is that it is integrable, that, if we have a force ${\vec F}$, then it is possible to find a potential $\phi({\vec x})$ such that ${\vec F} = - {\vec \nabla}\phi$. The reason for this is that if we pick out two points $p$ and $q$, we want the difference in energy between the two points to be $\phi(p) - \phi(q)$, and ...

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"Curl" is a pretty well named mathematical term--it denotes the degree of "rotation" in the vector field. For this reason, if you go all the way around in a vector field, you'll find that the total integral along that path will depend on the curl of the field in question. If a force had a curl, you could go all the way around and have some net work done, and ...

-2

The better way of looking at it would be from the macroscopic point of view.In a equipotential surface a particle that was under acceleration due to external force now no longer needs this external force to be applied. Hence the value of force become 0, resulting in zero work done...

2

$\vec F \cdot \mathrm{d} \vec r$ does not determine if a force is conservative or not. All normal forces (conservative or not) produce work equal to $\vec F \cdot \mathrm{d} \vec r$ but what determines if they are conservative is the integral of $\vec F \cdot \mathrm{d} \vec r$ in a closed loop. If that is equal to zero i.e. if \$\oint \vec F \cdot \mathrm{d} ...

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