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Work is done as long as the force is applied on the body, so in this case, the total work done would be the product of Force applied and the Displacement during the initial push only. If the object moves forever, it would do so with a constant velocity in this scenario, and consequentially its Kinetic Energy would be constant, implying that the work done ...


4

If the spring has been stretched then something must have stretched it. It's the work done by that something that you're calculating. That work has been done on the spring, so the potential energy of the spring has increased by an amount equal to the work done on it. To give a concrete example, suppose you're holding the end of the spring. Then your hand ...


4

"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 ...


3

The equation you need is that force is equal to the rate of change of momentum. The force is the weight of the rocket, $Mg$, and the rate of change of momentum is the mass ejected from the exhaust per second multiplied by the exhaust velocity. $$ Mg = v\frac{dm}{dt} $$ So choose your exhaust velocity $v$, and you can work out the required $dm/dt$. The ...


2

F.dr does not determine if a force is conservative or not.All normal forces (conservative or not) produce work equal to F.dr but what determines if they are conservative is the integral of F.dr in a closed loop.If that is equal to zero then it is conservative because no energy is lost in that loop.It is like gravity.You throw something upwards and no energy ...


2

At every point along the beam, the curvature has to be such that the externally applied bending moment exactly counters the internal stress. This tells you that the curvature is not constant - it is a function of distance to the side (largest in the middle, zero at the wall). This means that your assumption of "circular section" is wrong. See for example ...


2

Hint: $W=\vec{F}\cdot\vec{s}$ What components of $\vec{F}$ actually contribute to the work done? Hint 2: the dot product of two vectors $\vec{u}$ and $\vec{v}$ corresponds to the projection of one on the other, and is given by $\vec{u}\cdot\vec{v}=u_xv_x+u_yv_y+u_zv_z$. Can you find the $x$-, $y$- and $z$-components of $\vec{F}$ and $\vec{s}$? Hint 3: in ...


2

Really, the "lengths" you have chosen should represent positions (i.e. coordinates) in whatever space you're working in. The force integral is an integral over a path, with start and end points which become the limits of the integral. For a non-conservative force field, the integral will depend on the path chosen, so that too goes along with that integral. ...


1

If it is what you are saying then remember that work done here is neither negative nor positive however "work" doesn't exist here. We define Work as the product of the component of Net force acting on the body and the distance through which Force is exerted. You can say that the Force of gravity and Normal force are cancelling effects of each other since ...


1

The resolution to this problem is simple once you know how... Remember work done is force times distance moved in the direction of the force. The electrons are moving upwards, the Lorentz force $-ev \times B$ is in the direction shown in the diagram. BUT, the force did not do any work, because the force is perpendicular to the direction of travel of the ...


1

The formula you cite refers to the work done when the change of volume is made at constant pressure, which is not the case here. Let's say I start with the balloon in the air with volume $V_0$ at pressure $p_0$. The state I want to end up with is the balloon underwater at a depth with pressure $p_1$ and the balloon having volume $V_1$. I cannot directly ...


1

The work-energy principle is valid regardless of the presence of any non conservative forces. As long as you are using the work done by the resultant force (and resultant moment when involving rigid bodies) in the equation (or equivalently adding the work done by each force/moment), the work energy principle is valid. This can be shown by considering a ...


1

The difference is that $dW$ is an infinitesimal ''quantity'', whilst $W$ is not. I assume the context here is thermodynamics, which make use of calculus. In calculus there is the concept of the infinitesimal. I suggest, for you, to concern yourself with the structure of calculus if you are to tackle thermodynamics.


1

It depends on how you define the potential. In mechanics one has the convention $$ F = -\nabla U$$ Since the electric field exerts a force via $F=qE$ it is only natural to apply this convention in electrostatics too. In this way the electric potential $V$ can be directly interpreted as mechanical potential energy $U=qV$. Option number (2) is therefor the ...



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