91Ó°ÊÓ

• Mass of the box$\left(m\right)\mathrm{is}2.0\mathrm{kg}$and
• One of the forces$\left({\stackrel{â‡\P Ä}{F}}_1\right)is20\mathrm{N}$

Newton’s second law states that the net force$\stackrel{\mathbf{â‡\P Ä}}{\mathbf{F}}$ on a body with mass mis related to the body’s acceleration $\stackrel{\mathbf{â‡\P Ä}}{\mathbf{a}}$by,

$\stackrel{\mathbf{â‡\P Ä}}{\mathbf{F}}\mathbf{=}\mathit{m}\stackrel{\mathbf{â‡\P Ä}}{\mathbf{a}}$

Since there are two horizontal forces and the box is moving along the x-axis, we assume that the summation of all the forces equals the product of the mass of the box and the net acceleration produced.

Formulae:

$$\begin{gathered} F=ma\left(\mathrm{i}\right) \\ W=mg\left(\mathrm{ii}\right) \\ {\stackrel{â‡\P Ä}{F}}_1+{\stackrel{â‡\P Ä}{F}}_2=m\stackrel{â‡\P Ä}{a}\left(\mathrm{iii}\right) \end{gathered}$$

Let’s take the rightward direction as positive x direction.

When we plug the values in the equation (iii), we get,

$$\begin{gathered} 20\mathrm{N}+{\stackrel{â‡\P Ä}{\mathrm{F}}}_2=2.0\mathrm{kg}\left(10\mathrm{m}/{\mathrm{s}}^2\right) \\ {\stackrel{â‡\P Ä}{\mathrm{F}}}_2=0\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Therefore, the second force on the mass is $0\mathrm{m}/{\mathrm{s}}^2$.

Substituting the value of acceleration in equation (iii), we get,

$$\begin{gathered} 20\mathrm{N}+{\stackrel{â‡\P Ä}{\mathrm{F}}}_2=2.0\mathrm{kg}\left(20\mathrm{m}/{\mathrm{s}}^2\right) \\ {\stackrel{â‡\P Ä}{\mathrm{F}}}_2=\left(20\mathrm{N}\right)\hat{\mathrm{i}} \end{gathered}$$

Therefore, the second force on the mass is $\left(20\mathrm{N}\right)\hat{\mathrm{i}}$.

Substituting the value of acceleration in equation (iii), we get,

$$\begin{gathered} 20\mathrm{N}+{\stackrel{â‡\P Ä}{\mathrm{F}}}_2=2.0\mathrm{kg}\left(0\mathrm{m}/{\mathrm{s}}^2\right) \\ {\stackrel{â‡\P Ä}{\mathrm{F}}}_2=\left(-20\mathrm{N}\right)\hat{\mathrm{i}} \end{gathered}$$

Therefore, the second force on the mass is $\left(-20\mathrm{N}\right)\hat{\mathrm{i}}$.

Substituting the value of acceleration in equation (iii), we get,

$$\begin{gathered} 20\mathrm{N}+{\stackrel{â‡\P Ä}{\mathrm{F}}}_2=2.0\mathrm{kg}\left(-10\mathrm{m}/{\mathrm{s}}^2\right) \\ {\stackrel{â‡\P Ä}{\mathrm{F}}}_2=\left(-40\mathrm{N}\right)\hat{\mathrm{i}} \end{gathered}$$

Therefore, the second force on the mass is $\left(-40\mathrm{N}\right)\hat{\mathrm{i}}$.

Substituting the value of acceleration in equation (iii), we get

$$\begin{gathered} 20\mathrm{N}+{\stackrel{â‡\P Ä}{\mathrm{F}}}_2=2.0\mathrm{kg}\left(-20\mathrm{m}/{\mathrm{s}}^2\right) \\ {\stackrel{â‡\P Ä}{\mathrm{F}}}_2=\left(-60\mathrm{N}\right)\hat{\mathrm{i}} \end{gathered}$$

Therefore, the second force on the mass is $\left(-60\mathrm{N}\right)\hat{\mathrm{i}}$.