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Write Newton's Second Law of Motion in vector form.

Short Answer

Expert verified
Answer: Newton's Second Law of Motion in vector form is given by the equation \(\vec{F} = m \cdot \vec{a}\), where \(\vec{F}\) represents the force vector, m represents the mass, and \(\vec{a}\) represents the acceleration vector.

Step by step solution

01

Recall Newton's Second Law of Motion

Newton's Second Law of Motion states that the force acting on an object is equal to the mass of the object multiplied by its acceleration. It is given by the equation F = m * a, where F represents the force, m represents the mass, and a represents the acceleration of the object.
02

Represent force, mass, and acceleration as vectors

In order to write Newton's Second Law in vector form, we need to express force, mass, and acceleration as vectors. The force and acceleration are vector quantities because they have both magnitude and direction. The mass is a scalar quantity since it has only magnitude. A vector is denoted as a bold letter or with an arrow on top of the symbol. Thus, force vector is represented as **F** or \(\vec{F}\), and acceleration vector is represented as **a** or \(\vec{a}\).
03

Write Newton's Second Law in vector form

We can now write Newton's Second Law of Motion in its vector form by replacing the force and acceleration scalar symbols with their vector representations: \(\vec{F} = m \cdot \vec{a}\) This equation represents Newton's Second Law of Motion in vector form, which states that the force vector acting on an object is equal to the mass of the object multiplied by its acceleration vector.

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Most popular questions from this chapter

Orthogonal lines Recall that two lines \(y=m x+b\) and \(y=n x+c\) are orthogonal provided \(m n=-1\) (the slopes are negative reciprocals of each other). Prove that the condition \(m n=-1\) is equivalent to the orthogonality condition \(\mathbf{u} \cdot \mathbf{v}=0,\) where \(\mathbf{u}\) points in the direction of one line and \(\mathbf{v}\) points in the direction of the other line..

An object moves along a path given by $$\mathbf{r}(t)=\langle a \cos t+b \sin t, c \cos t+d \sin t, e \cos t+f \sin t\rangle$$ for \(0 \leq t \leq 2 \pi\) a. Show that the curve described by \(\mathbf{r}\) lies in a plane. b. What conditions on \(a, b, c, d, e,\) and \(f\) guarantee that the curve described by \(\mathbf{r}\) is a circle?

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Find the function \(\mathbf{r}\) that satisfies the given conditions. $$\mathbf{r}^{\prime}(t)=\left\langle e^{2 t}, 1-2 e^{-t}, 1-2 e^{t}\right\rangle ; \mathbf{r}(0)=\langle 1,1,1\rangle$$

Compute the following derivatives. $$\frac{d}{d t}\left(\left(3 t^{2} \mathbf{i}+\sqrt{t} \mathbf{j}-2 t^{-1} \mathbf{k}\right) \cdot(\cos t \mathbf{i}+\sin 2 t \mathbf{j}-3 t \mathbf{k})\right)$$

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