Question

There are two horizontal forces on the 2.0 kgbox in the overhead view of the Figure but only one (of magnitude${\mathbf{F}}_{\mathbf{1}}\mathbf{=}\mathbf{20}\mathbf{}\mathbf{N}$) is shown. The box moves along the xaxis. For each of the following values for the acceleration${\mathit{a}}_{\mathbf{x}}$of the box, find the second force in unit-vector notation$\mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{}\mathbf{10}\mathbf{}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}\mathbf{}\mathbf{\left(}\mathbf{b}\mathbf{\right)}\mathbf{}\mathbf{20}\mathbf{}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}\mathbf{}\mathbf{\left(}\mathbf{c}\mathbf{\right)}\mathbf{}\mathbf{0}\mathbf{}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}\mathbf{}\mathbf{\left(}\mathbf{d}\mathbf{\right)}\mathbf{-}\mathbf{10}\mathbf{}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}\mathbf{}\mathbf{\left(}\mathbf{e}\mathbf{\right)}\mathbf{-}\mathbf{20}\mathbf{}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$

Short answer

(a) Second Force on the box with acceleration $10\mathrm{m}/{\mathrm{s}}^2$is zero.

(b) Second Force on the box with acceleration $20\mathrm{m}/{\mathrm{s}}^2\mathrm{is}\left(20\mathrm{N}\right)\hat{\mathrm{i}}$.

(c) Second Force on the box with acceleration $0\mathrm{m}/{\mathrm{s}}^2\mathrm{is}\left(-20\mathrm{N}\right)\hat{\mathrm{i}}$.

(d) Second Force on the box with acceleration $20\mathrm{m}/{\mathrm{s}}^2\mathrm{is}\left(-10\mathrm{N}\right)\hat{\mathrm{i}}$.

(e) Second Force on the box with acceleration $20\mathrm{m}/{\mathrm{s}}^2\mathrm{is}\left(-60\mathrm{N}\right)\hat{\mathrm{i}}$.

Step-by-step solution

Given data

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

Understanding the concept of the net force

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

$\stackrel{\mathbf{\rightharpoonup }}{\mathbf{F}}\mathbf{=}\mathit{m}\stackrel{\mathbf{\rightharpoonup }}{\mathbf{a}}$

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:

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

(a) Find the second force in unit-vector notation when  a=(10m/s2)

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

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

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

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

(b) Find the second force in unit-vector notation when a=(20 m/s2) 

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

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

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

(c) Find the second force in unit-vector notation when a=(0 m/s2)

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

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

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

(d) Find the second force in unit-vector notation when a=(-10m/s2)

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

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

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

(e) Find the second force in unit-vector notation when a=(-20 m/s2)

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

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

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