📂Classical Mechanics

Day, Work-Energy Theorem

Definition

A work is defined as the product of the magnitude of a force $\mathbf{F}$ and the distance $s$ by which an object has moved in the same direction as the force $W=Fs$, when the force $\mathbf{F}$ acts on the object.

Description

In the distance-force graph, the area under the graph is equal to the amount of work. Since the direction of movement and the direction of the force must be the same for work to be done on the object, it can be expressed as the dot product of the two vectors.

$$ W= \mathbf{F} \cdot \mathbf{s}=Fs\cos \theta $$

Since it was said to be equal to the area of the distance-force graph, it can also be expressed as follows.

$$ W=\int _{s_{0}}^{s} Fds $$

Why it is said that work is done only when the movement direction and the force direction are the same can be easily understood intuitively. See the picture below.

The picture above shows the situation where $A$ and $B$ are pushing the object in the direction of the arrows. If the object moved in the direction of the green arrow, we consider that $B$ did not play a part in moving the object. It was moved solely by $A$. Therefore, when calculating the work done by a force on an object, only the cases where the force and the direction of movement are the same are considered. If the direction of movement and the direction of the force are exactly opposite, it is considered that the force does work negatively on the object. For example, consider the case where force is applied in the opposite direction to stop a moving object as shown in the picture below.

Work-Energy Theorem¹

The net work done by a force on an object is equal to the change in kinetic energy of the object.

$$ W=\Delta T $$

Proof

By the definition of kinetic energy, it is $F(x)=\frac{dT}{dx}$, hence it can be expressed as follows.

$$ W=\int_{x_{0}}^{x}Fdx=\int_{x_{0}}^{x}dT=T-T_{0}=\Delta T $$

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