Chapter 5, Problem 23PQ

The x and y coordinates of a 4.00-kg particle moving in the xy plane under the influence of a net force F are given by x = t⁴ − 6t and y = 4t² + 1, with x and y in meters and t in seconds. What is the magnitude of the force F at t = 4.00 s?

To determine

Find the magnitude of the force at $t=4.00\text{ s}$.

Answer to Problem 23PQ

The magnitude of force at $t=4.00\text{ s}$ is $\text{769 N}$.

Explanation of Solution

Write the Newton’s second law of motion for x and y component.

  $\begin{array}{l}{F}_x=m{a}_x\\ F_y=m{a}_y\end{array}$                                                                                              (I)

Here, $F_x$ is the x component of force of the particle, $a_x$ is the x component of acceleration, $F_y$ is the y component of force of the particle, $a_y$ is the y component of acceleration and $m$ is the mass of the particle.

Write the magnitude of the force.

    $F_{net}=\sqrt{F_x^2+F_y^2}$                                                                                              (II)

Here, $F_{net}$ is the magnitude of the force.

Write the x and y component of the velocity and acceleration of the particle.

    $\begin{array}{l}{v}_x=\frac{dx}{dt}\\ v_y=\frac{dy}{dt}\end{array}$                                                                                               (III)

  $\begin{array}{l}{a}_x=\frac{d{v}_x}{dt}\\ a_y=\frac{d{v}_y}{dt}\end{array}$                                                                                             (IV)

Here, $v_x$ and $v_y$ are x and y component of velocity and $a_x$ and $a_y$ are x and y component of acceleration.

Conclusion:

Substitute $t^4-6t$ for $x$, $4t^2+1$ for $y$ in equation III.

    $\begin{array}{c}{v}_x=\frac{d\left(t^4-6t\right)}{dt}\\ =4t^3-6\end{array}$

    $\begin{array}{c}{v}_y=\frac{d\left(4t^2+1\right)}{dt}\\ =8t\end{array}$

Substitute $4t^3-6$ for $v_x$ and $8t$ for $v_y$ in equation IV.

    $\begin{array}{c}{a}_x=\frac{d\left(4t^3-6\right)}{dt}\\ =12t^2\end{array}$

    $\begin{array}{c}{a}_y=\frac{d\left(8t\right)}{dt}\\ =8{\text{ m/s}}^2\end{array}$

Substitute $4.00\text{ kg}$ for $m$, $12t^2$ for $a_x$, $8{\text{ m/s}}^2$ for $a_y$ in equation I.

    $\begin{array}{c}{F}_x=\left(4.00\text{ kg}\right)\left(12t^2\right)\\ =48.0t^2\text{ N}\end{array}$

    $\begin{array}{c}{F}_y=\left(4.00\text{ kg}\right)\left(8{\text{ m/s}}^2\right)\\ =32.0\text{ N}\end{array}$

Substitute $48.0t^2\text{ N}$ for $F_x$, $4.00\text{ s}$ for $t$ and $32.0\text{ N}$ for $F_y$ in equation II.

    $\begin{array}{c}{F}_{net}=\sqrt{{\left[\left(48.0\right){\left(4.00\text{ s}\right)}^2\right]}^2+{\left(32.0\text{ N}\right)}^2}\\ =769\text{ N}\end{array}$

Therefore, the magnitude of force at $t=4.00\text{ s}$ is $\text{769 N}$.