Chapter 7, Problem 7.11P

A force $\stackrel{\to }{\text{F}}$ = (6 $\hat{\text{i}}$ − 2 $\hat{\text{j}}$) N acts on a panicle that under-goes a displacement Δ $\stackrel{\to }{\text{r}}$ = (3 $\hat{\text{i}}$ + $\hat{\text{j}}$) m. Find (a) the work done by the force on the particle and (b) the angle between $\stackrel{\to }{\text{F}}$ and Δ $\stackrel{\to }{\text{r}}$.

To determine

The work done on the particle by the force.

Answer to Problem 7.11P

The work done on the particle is $16.\text{0}\text{J}$.

Explanation of Solution

Given info: A force vector is $\left(6\widehat{\text{i}}-2\widehat{\text{j}}\right)\text{N}$ and the displacement vector is $\left(3\widehat{\text{i}}+\widehat{\text{j}}\right)\text{m}$.

From the properties of dot products, the relation of unit like vectors is,

$\widehat{\text{i}}\cdot \widehat{\text{i}}=\widehat{\text{j}}\cdot \widehat{\text{j}}=\widehat{\text{k}}\cdot \widehat{\text{k}}=1$

From the properties of dot products, the relation of unit unlike vectors is,

$\widehat{\text{i}}\cdot \widehat{\text{j}}=\widehat{\text{j}}\cdot \widehat{\text{k}}=\widehat{\text{k}}\cdot \widehat{\text{i}}=0$

The expression for the work done on the particle is,

Here,

is the force vector.

is the displacement vector.

Substitute $\left(6\widehat{\text{i}}-2\widehat{\text{j}}\right)\text{N}$ for and $\left(3\widehat{\text{i}}+\widehat{\text{j}}\right)\text{m}$ for in the above equation.

$\begin{array}{c}W=\left(6\widehat{\text{i}}-2\widehat{\text{j}}\right)\text{N}\cdot \left(3\widehat{\text{i}}+\widehat{\text{j}}\right)\text{m}\\ =\left[\left(6\widehat{\text{i}}\right)\cdot \left(3\widehat{\text{i}}\right)+\left(6\widehat{\text{i}}\right)\cdot \left(\widehat{\text{j}}\right)-\left(2\widehat{\text{j}}\right)\cdot \left(3\widehat{\text{i}}\right)-\left(2\widehat{\text{j}}\right)\cdot \left(\widehat{\text{j}}\right)\right]\text{N}\cdot \text{m}\\ =\left[\left(18\right)\left(\widehat{\text{i}}\cdot \widehat{\text{i}}\right)+\left(6\right)\left(\widehat{\text{i}}\cdot \widehat{\text{j}}\right)-\left(\text{6}\right)\left(\widehat{\text{j}}\cdot \widehat{\text{i}}\right)-\left(2\right)\left(\widehat{\text{j}}\cdot \widehat{\text{j}}\right)\right]\text{N}\cdot \text{m}\end{array}$

Substitute $1$ for $\widehat{\text{i}}\cdot \widehat{\text{i}}$, $1$ for $\widehat{\text{j}}\cdot \widehat{\text{j}}$, and $0$ for $\widehat{\text{i}}\cdot \widehat{\text{j}}$ in the above equation.

$\begin{array}{c}W=\left[\left(18\right)\left(1\right)+\left(6\right)\left(0\right)-\left(\text{6}\right)\left(0\right)-\left(2\right)\left(1\right)\right]\text{N}\cdot \text{m}\\ =16.\text{0}\text{J}\end{array}$

Conclusion:

Therefore, the work done on the particle is $16.\text{0}\text{J}$.

To determine

The angle between and .

Answer to Problem 7.11P

The angle between and is $36.9°$.

Explanation of Solution

From part (a), the work done on the particle by the force is,

$W=16.\text{0}\text{J}$

The equation of work done on the particle is,

(1)

Here,

is the magnitude of vector force.

is the magnitude of displacement vector.

$\theta$ is the angle between and .

The force vector is expressed as,

Here,

$F_x$ is the $x$-component of the force.

$F_y$ is the $y$-component of the force.

$F_z$ is the $z$-component of the force.

Compare the given force vector with the above equation, the value of $F_x$ is $6\text{N}$, $F_y$ is $-2\text{N}$ and $F_z$ is $0$.

The magnitude of force vector is,

Substitute $6\text{N}$ for $F_x$, $-2\text{N}$ for $F_y$ and $0$ for $F_z$ in the above equation.

The displacement vector is expressed as,

$\Delta \text{r}=r_x\widehat{\text{i}}+r_y\widehat{\text{j}}+r_z\widehat{\text{k}}$

Here,

$r_x$ is the $x$-component of the displacement.

$r_y$ is the $y$-component of the displacement.

$r_z$ is the $z$-component of the displacement.

Compare the given displacement vector with the above equation, the value of $r_x$ is $3\text{m}$, $r_y$ is $1\text{m}$ and $r_z$ is $0$.

The magnitude of displacement vector is,

Substitute $3\text{m}$ for $r_x$, $1\text{m}$ for $r_y$ and $0$ for $r_z$ in the above equation.

Substitute $\text{16}\text{J}$ for $W$, $6.325\text{N}$ for and $3.162\text{m}$ for in equation (1).

$\begin{array}{c}\text{16}\text{J}=\left(6.32\text{N}\right)\left(3.16\text{m}\right)\cos \theta \\ \theta ={\cos }^{-1}\left(\frac{\text{16}\text{J}}{\left(6.325\text{N}\right)\left(3.162\text{m}\right)}\right)\\ =36.8698°\\ \approx 36.9°\end{array}$

Conclusion:

Therefore, the angle between and is $36.9°$.