Question

Two horizontal forces act on a 2.0kgchopping block that can slide over a frictionless kitchen counter, which lies in an x-yplane. One force is ${\overrightarrow{\mathbf{F}}}_{\mathbf{1}}\mathbf{=}\left(3N\right)\stackrel{\mathbf{⏜}}{\mathbf{i}}\mathbf{+}\left(4N\right)\stackrel{\mathbf{⏜}}{\mathbf{J}}$. Find the acceleration of the chopping block in unit-vector notation when the other force is (a) role="math" localid="1657018090784" ${\overrightarrow{\mathbf{F}}}_{\mathbf{2}}\mathbf{=}\mathbf{(}\mathbf{-}\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{N}\mathbf{)}\stackrel{\mathbf{⏜}}{\mathbf{i}}\mathbf{+}\mathbf{(}\mathbf{-}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{N}\mathbf{)}\stackrel{\mathbf{⏜}}{\mathbf{J}}$(b) Find the acceleration of the chopping block in unit-vector notation when the other force is role="math" localid="1657018141943" ${\overrightarrow{\mathbf{F}}}_{\mathbf{2}}\mathbf{=}\mathbf{(}\mathbf{-}\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{N}\mathbf{)}\stackrel{\mathbf{⏜}}{\mathbf{i}}\mathbf{+}\mathbf{(}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{N}\mathbf{)}\stackrel{\mathbf{⏜}}{\mathbf{J}}$ and (c) Find the acceleration of the chopping block in unit-vector notation when the other force is ${\overrightarrow{\mathbf{F}}}_{\mathbf{2}}\mathbf{=}\mathbf{(}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{N}\mathbf{)}\stackrel{\mathbf{⏜}}{\mathbf{i}}\mathbf{+}\mathbf{(}\mathbf{-}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{N}\mathbf{)}\stackrel{\mathbf{⏜}}{\mathbf{J}}$

Short answer

The acceleration of the chopping block in unit vector notation when

a) ${\overrightarrow{F}}_2=\left(-3.0\mathrm{N}\right)\stackrel{⏜}{i}+\left(-4.0N\right)\stackrel{⏜}{j}$is $0\mathrm{m}/{\mathrm{s}}^2$.

b) role="math" localid="1657018476910" ${\overrightarrow{F}}_2=\left(-3.0N\right)\stackrel{⏜}{i}+\left(4.0N\right)\stackrel{⏜}{j}is4.0\stackrel{⏜}{j}\mathrm{m}/{\mathrm{s}}^2$.

c)${\overrightarrow{F}}_2=\left(-3.0N\right)\stackrel{⏜}{i}+\left(-4.0N\right)\stackrel{⏜}{j}is3.0\stackrel{⏜}{j}\mathrm{m}/{\mathrm{s}}^2$.

Step-by-step solution

The given data

• Mass of chopping box, $m=2.0kg$
• First force in vector form, ${\overrightarrow{F}}_1=\left(3.0N\right)\stackrel{⏜}{i}+\left(4.0N\right)\stackrel{⏜}{j}$
• Three other different forces,

$$\begin{gathered} {\overrightarrow{F}}_2=\left(-3.0N\right)\stackrel{⏜}{i}+\left(-4.0N\right)\stackrel{⏜}{j} \\ {\overrightarrow{F}}_2=\left(-3.0N\right)\stackrel{⏜}{i}+\left(4.0N\right)\stackrel{⏜}{j} \\ {\overrightarrow{F}}_2=\left(3.0N\right)\stackrel{⏜}{i}+\left(-4.0N\right)\stackrel{⏜}{j} \end{gathered}$$

Understanding the concept of Newton’s law of motion

From Newton’s second law of motion, the force acting on the body is equal to the product of mass and its acceleration. The net force is the vector sum of all the forces acting on the body. The net force ${\overrightarrow{\mathbf{F}}}_{\mathbf{N}\mathbf{e}\mathbf{t}}$on a body with a mass m and acceleration $\overrightarrow{\mathbf{a}}$is given by,

${\overrightarrow{\mathbf{F}}}_{\mathbf{N}\mathbf{e}\mathbf{t}}\mathbf{=}\mathit{m}\overrightarrow{\mathbf{a}}$

Using the formula for force from Newton’s second law, we can find the acceleration of the chopping block in unit vector notation.

Formulae:

The net force on a particle according to Newton’s second law,

$$\begin{gathered} {\overrightarrow{F}}_{net}={\overrightarrow{F}}_1+{\overrightarrow{F}}_2\left(i\right) \\ =\mathrm{M}\overrightarrow{\mathrm{a}} \end{gathered}$$

(a) Calculations for the magnitude of acceleration when F→2=(-3.0N)i⏜+(-4.0N)j⏜

From equation (i), we get the value of acceleration of the body. Substitute the values of forces and mass in the equation (i).

$$\begin{gathered} \overrightarrow{a}=\frac{\left({\overrightarrow{F}}_1+{\overrightarrow{F}}_2\right)}{m} \\ =\frac{\left[\left(3.0N\right)\stackrel{⏜}{i}+\left(4.0N\right)\stackrel{⏜}{j}+\left(-3.0N\right)\stackrel{⏜}{i}+\left(-4.0N\right)\stackrel{⏜}{j}\right]}{2.00\mathrm{kg}} \\ =0\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Hence, the value of acceleration is $0\mathrm{m}/{\mathrm{s}}^2$.

(b) Calculations for the magnitude of acceleration when F→2=(-3.0N)i⏜+(4.0N)j⏜ 

From equation (i), we get the value of acceleration of the body. Substitute the values of forces and mass in the equation (i).

$$\begin{gathered} \overrightarrow{a}=\frac{\left({\overrightarrow{F}}_1+{\overrightarrow{F}}_2\right)}{m} \\ =\frac{\left[\left(3.0\mathrm{N}\right)\stackrel{⏜}{i}+\left(4.0\mathrm{N}\right)\stackrel{⏜}{j}+\left(-3.0\mathrm{N}\right)\stackrel{⏜}{i}+\left(4.0\mathrm{N}\right)\stackrel{⏜}{j}\right]}{2.00\mathrm{kg}} \\ =4.0\stackrel{⏜}{j}\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Hence, the value of acceleration is $4.0\stackrel{⏜}{j}\mathrm{m}/{\mathrm{s}}^2$.

(c) Calculations for the magnitude of acceleration when F→2=(3.0N)i⏜+(-4.0N)j⏜

From equation (i), we get the value of acceleration of the body. Substitute the values of forces and mass in the equation (i).

$$\begin{gathered} \overrightarrow{a}=\frac{\left({\overrightarrow{F}}_1+{\overrightarrow{F}}_2\right)}{m} \\ =\frac{\left[\left(3.0\mathrm{N}\right)\stackrel{⏜}{i}+\left(4.0\mathrm{N}\right)\stackrel{⏜}{j}+\left(+3.0\mathrm{N}\right)\stackrel{⏜}{i}+\left(4.0\mathrm{N}\right)\stackrel{⏜}{j}\right]}{2.00\mathrm{kg}} \\ =3.0\stackrel{⏜}{i}\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Hence, the value of acceleration is$3.0\stackrel{⏜}{i}\mathrm{m}/{\mathrm{s}}^2$ .