Find the magnitude of the net force on an object of mass 10 kg with the given position function

Name: Problem 15, Chapter 11: Calculus: Early Transcendental Functions
Uploaded: 2022-10-08T02:42:38-08:00
Duration: 3 min 27 s
Channel: Steven Clarke
Description: Find the magnitude of the net force on an object of mass 10 kg with the given position function (in units of meters and seconds). 𝐫(t)=⟨4 cos2 t, 4 sin2 t⟩

step 1 Okay, I want to find the force acting on this object, given R. Now, that means I want to find ac

Question

Find the magnitude of the net force on an object of mass $10 \mathrm{~kg}$ with the given position function (in units of meters and seconds). $$ \mathbf{r}(t)=\langle 4 \cos 2 t, 4 \sin 2 t\rangle $$

   Find the magnitude of the net force on an object of mass $10 \mathrm{~kg}$ with the given position function (in units of meters and seconds).
$$
\mathbf{r}(t)=\langle 4 \cos 2 t, 4 \sin 2 t\rangle
$$

Calculus: Early Transcendental Functions

Robert Smith, Roland… 4th Edition

Chapter 11, Problem 15

Instant Answer

Solved by Expert Steven Clarke

10/08/2022

Step 1

The velocity is the derivative of the position function with respect to time. So, we differentiate $\mathbf{r}(t)=\langle 4 \cos 2 t, 4 \sin 2 t\rangle$ with respect to $t$ to get the velocity $\mathbf{v}(t)$: $$ \mathbf{v}(t) = \frac{d\mathbf{r}}{dt} = \langle -8 Show more…

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Find the magnitude of the net force on an object of mass $10 \mathrm{~kg}$ with the given position function (in units of meters and seconds). $$ \mathbf{r}(t)=\langle 4 \cos 2 t, 4 \sin 2 t\rangle $$

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Key Concepts

Newton's Second Law

Newton's second law of motion states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. This principle is crucial for linking kinematic descriptions of motion to the forces responsible for that motion.

Uniform Circular Motion

Uniform circular motion describes the motion of an object traveling at a constant speed along a circular path. Even though the speed is constant, the object's velocity vector changes direction continuously, resulting in a centripetal acceleration directed toward the center of the circle.

Derivatives in Kinematics

Taking the derivative of the position vector with respect to time yields the velocity vector, and a second derivative gives the acceleration vector. These processes are essential in analyzing the dynamical properties of an object's motion and determining forces acting on it.

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