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Due to an accident of history there are two different units called the calorie and the Calorie - yes, the only difference is the capital C. The calorie is 4.2J but the Calorie is 4.2kJ, and the calories counted in diets are actually Calories even though they are invariably written on the food packaging with a small c. So you only used 3 Calories walking up ...

4

1&2) Let's suppose there are only two forces in play: the one exerted by the man, and weight. The sum of works done on the body equals the variation of kinetic energy (this is a theorem of mechanics; I fail to find its English name). Since the body has no velocity at the start and at the end, the sum of works is zero. The net work done by the man on ...

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Walking requires raising and lowering the centre of gravity, as well as moving the limbs. Both of these require muscles to contract and extend and use energy. If the muscles were perfect springs then the energy stored during contraction could be fully recovered during extension. No energy would be lost. But they are not perfect springs - some energy is ...

3

There is a little bit more thinking behind saying that $P=\vec F \cdot \vec v$ than it being a generalised multiplication in 3D. There are even cases where multiplication with scalar becomes a cross product when using 3D vectors. For example, torque $T=Fr$, becomes $\vec T = \vec r \times \vec F$. Whenever implementing vectors into existing scalar equations, ...

2

To my understanding, work is done on object A when object B is applying a force on object A, causing object A to displace. Work is done whenever a force displaces an object. Since energy is the ability to do work, what work does a moving object do, due to its kinetic energy? A moving object might not do any work at all. Imagine an empty ...

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You should plug in the horizontal force acting on the object if you want to know the work done by this force. Due to friction heating up the plane not all of this work is converted in the objects kinetic energy. PS: I assume you meant "while calculating the work" rather than "while calculating the force"..

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A force field is called conservative if its work between any points $A$ and $B$ does not depend on the path. This implies that the work over any closed path (circulation) is zero. This also implies that the force cannot depend explicitly on time. Consider for instance a time decaying force on a straight line. Choose a long closed path. The magnitude of the ...

2

Are isothermal free expansion and adiabatic free expansion different? No. They are the same. Your mistake is in thinking that $PV^\gamma = \text{constant}$ applies to a free expansion. That expression is for a reversible (i.e., isentropic) adiabatic process. A gas that has undergone a free expansion has more entropy after the expansion is complete than ...

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1) What can be say about the work done by the man to the weight? Unlike gravity, the force exerted by the man is in general not constant, does not depend only on position of the body (you may apply different force on the weight at the same height $h$ on the way up vs. on the way down), and is not conservative (does not arise from a potential). When the ...

2

The collision doesn't happen at a single point in space - rather the colliding objects exert a forces on each other over a distance as they approach and the recede. Consider a tennis ball hitting a racket - the ball and the strings of the racket deform and we get an increasing elastic restoring forces until the two objects at at their closest approach. ...

2

From the definition of work $$W = \int dx F$$ and $$P = \frac{dW}{dt}$$ you can see how we can arrive at $$P = F \frac{dx}{dt} = F v$$ (when considering only the absolute value). To understand it intuitively, imagine the case of a frictionless system in which the car can move at a certain speed without any opposing force. The power required to keep it at ...

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Assuming no air friction, you compute the initial velocity needed from conservation of energy: $$\frac12 mv^2=mgh$$ The impulse needed is $mv=F\Delta t$. The product of these ($F, \Delta t$) is constant - shorter time implies higher force. The above assumes the time of impact is short enough not to affect the over all time (otherwise you need to solve ...

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To understand why holding objects costs energy even though the work appears to be zero, you have to understand how muscles work. When you are holding an object, your muscles are contracted. The process of muscle contraction consists in a protein filament called Myosin pulling another filament, called Actin. Since this is a dynamical process (the Actin ...

1

The correct statement is - the total work done on the weight is zero i.e. the total energy of the weight before and after the experiment is same. However, when the man is lifting the weight he is obviously working against gravity. More importantly, when he is lowering the weight, he is still working against gravity, as gravity would rather lower the weight ...

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Work (or energy) is transferred from one particle to another, but the net effect is no change overall. How? Consider a collision force acting between two particles over a small time frame. During that time frame the particles move, and the work done on one particle is ${\rm d}W_1 = \boldsymbol{F} \cdot {\rm d}\boldsymbol{x}_1$. Since an equal and opposite ...

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As you know, work is calculated by force product distance that parallel to force. So the work that you want calculates below:

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The reason is that work always depends on the actual velocity the body has during its movement, even if you compute the work for only one force out of many. In other words, $W_{F_i} = \int F_i\cdot v dt$, rather than, as you might perhaps expect, $W_{F_i} = \int F_i \cdot v_i dt$, where $v_i$ is a hypothetical "what $v$ would be if only $F_i$ acted". Now ...

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John Rennie has explained the problems with units here. Now, you'll burn about a factor 4 more than the work you perform, due to losses when glucose or fats are burned to allow the muscles to do the work. The Gibbs free energy change when glucose or fat reacts with oxygen and changes into water and carbon dioxide gives you the maximum amount of work that can ...

1

You are not defining your systems clearly enough. If the system is the mass and the Earth then in the situation that you have described the external force doing work on that system increases the potential energy of the mass-Earth system. Now look at the system which is the mass alone. Then there are two forces acting on the mass. The force as described ...

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By work-KE theorem, $$\Delta \text{KE} = \text{Work done by NET force}$$ $$=\text{Work done by} F_1+\text{Work done by} F_2+\cdots$$ Now if a certain force $F_i$ is conservative, you have the choice of defining its corresponding PE so that $$\text{Work done by}F_i=-\Delta \text{PE}_i$$ and MOVE it to the left hand side so that you have ...

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The equation $\Delta U = Q - W$ is complete in itself. The confusion arises in the definition of enthalpy. We tend to think that the "pressure energy" $PV$ and the work done by the system $W$ are somehow different. But the fact is that there isn't a well defined physical meaning to the $PV$ term. It is confusing because when we add $U$ and $PV$, we think ...

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The javelin does work on the athlete's arm/hand equal in magnitude and opposite in sign to the work that the athlete's arm does on the javelin. Both the arm and the javelin experience a force applied through a displacement.

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The potential energy is the ability to perform work in the way that... potential energy $U$ is converted into kinetic $K$ energy during the fall. The kinetic energy implies a speed $v$ and thus causes a momentum $p=mv$. On impact this momentum is reduced drastically causing a force $F=dp/dt$. Is the impact-surface displaced by x, then the work $W=Fx$ has ...

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It's true that there are many inner products you can choose on $\mathbb{R}^3$. However, physics supplies the additional principle of rotational invariance: the result should not depend on our coordinate system. Now, any inner product of vectors $a$ and $b$ can be written as $$a \cdot b = a^T M b$$ for a matrix $M$. Rotational invariance tells us that $M$ ...

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