91影视

Mass of the suitcase is m = 20.0 kg

Inclined of the ramp is$胃=32掳$

The magnitude of the force is F = 160 N

The coefficient of kinetic friction is ${\mathrm{渭}}_{\mathrm{k}}=0.300$

The distance along the ramp is s = 3.80 m

The free-body diagram for the forces on the suitcase is shown in the below figure.

The expression for the work done in terms of force and displacement is,

$$\begin{gathered} \mathbf{W}\mathbf{=}\stackrel{\mathbf{鈫�}}{\mathbf{F}}\mathbf{路}\stackrel{\mathbf{鈫�}}{\mathbf{d}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{贵诲肠辞蝉蠒} \end{gathered}$$鈥︹赌︹赌︹赌︹赌︹赌�.(1)

Here, $\stackrel{\mathbf{鈫�}}{\mathbf{F}}$is the force exerted on the object, $\stackrel{\mathbf{鈫�}}{\mathbf{d}}$is the distance covered by the object, and$\mathbf{胃}$ is the angle between force and displacement.

Work done by the force on the suitcase is given by the equation (1),

Since the force exerted on the suitcase and the displacement of the suitcase are in the same direction then $\mathrm{蠒}=0掳$.

Substitute the given values in equation (1), and we get

$$\begin{gathered} {\mathrm{W}}_{\mathrm{F}}=\mathrm{贵蝉肠辞蝉蠒} \\ =\left(160\mathrm{N}\right)\left(3.8\mathrm{m}\right)\left(\cos 0掳\right) \\ =680\mathrm{J} \end{gathered}$$

Hence, the work done on the suitcase by the force is $680\mathrm{J}$.

Force due to gravity,

${\mathrm{F}}_{\mathrm{g}}=\mathrm{mg}$and

$$\begin{gathered} \mathrm{蠒}=90掳+32掳 \\ =122掳 \end{gathered}$$

Thus, the work done according to equation (1) is

${\mathrm{W}}_{\mathrm{g}}=\left(\mathrm{mg}\right)\mathrm{蝉肠辞蝉蠒}$

Substituting thegiven data in the above expression, we get,

$$\begin{gathered} {\mathrm{W}}_{\mathrm{g}}=\left(20\mathrm{kg}\right)\left(9.8\mathrm{m}路{\mathrm{s}}^{-2}\right)\left(3.8\mathrm{m}\right)\left(\cos 122掳\right) \\ =-395\mathrm{J} \end{gathered}$$

Hence, the work on the suitcase by gravity is $-395\mathrm{J}$.

Since the normal force ${\mathrm{F}}_{\mathrm{N}}$is equal to role="math" localid="1664867732493" $\mathrm{尘驳肠辞蝉胃}$.

Therefore using equation (1), the work done can be written as,

role="math" localid="1664867805316" ${\mathrm{W}}_{\mathrm{N}}=\left(\mathrm{尘驳肠辞蝉胃}\right)\mathrm{蝉肠辞蝉蠒}$

Substituting the given data in the above expression, we get,

$$\begin{gathered} {\mathrm{W}}_{\mathrm{N}}=\left(\mathrm{尘驳肠辞蝉胃}\right)\mathrm{蝉肠辞蝉蠒} \\ =\left(20\mathrm{kg}\right)\left(9.8\mathrm{m}路{\mathrm{s}}^{-2}\right)\left(\cos 32掳\right)\left(3.8\mathrm{m}\right)\left(\cos 90掳\right) \\ =0\mathrm{J} \end{gathered}$$

Hence, the work done on the suitcase by the normal force is $0\mathrm{J}$.

Work done on the suitcase by the friction can be expressed using equation (1), such that

${\mathrm{W}}_{{\mathrm{f}}_{\mathrm{k}}}={\mathrm{f}}_{\mathrm{k}}\mathrm{蝉肠辞蝉蠒}$

Since kinetic frictional force ${\mathrm{f}}_{\mathrm{k}}$is equal to ${\mathrm{渭}}_{\mathrm{k}}\mathrm{尘驳肠辞蝉胃}$. So, the work done is,

${\mathrm{W}}_{{\mathrm{f}}_{\mathrm{x}}}=\left({\mathrm{渭}}_{\mathrm{k}}\mathrm{尘驳肠辞蝉胃}\right)\mathrm{蝉肠辞蝉蠒}$

Substituting the given data in the above expression, we get,

role="math" localid="1664868547640" $$\begin{gathered} {\mathrm{W}}_{{\mathrm{f}}_{\mathrm{x}}}=\left({\mathrm{渭}}_{\mathrm{k}}\mathrm{尘驳肠辞蝉胃}\right)\mathrm{蝉肠辞蝉蠒} \\ =\left(0.3\right)\left(20\mathrm{kg}\right)\left(9.8\mathrm{m}路{\mathrm{s}}^{-2}\right)\left(\cos 32掳\right)\left(3.8\mathrm{m}\right)\left(\cos 180掳\right) \\ =-189\mathrm{J} \end{gathered}$$

Hence, the work done on the suitcase by the friction is $-189\mathrm{J}$.

The net work done on the package is

${\mathrm{W}}_{\mathrm{net}}={\mathrm{W}}_{\mathrm{F}}+{\mathrm{W}}_{\mathrm{g}}+{\mathrm{W}}_{\mathrm{N}}+{\mathrm{W}}_{{\mathrm{f}}_{\mathrm{x}}}$

Here, ${\mathrm{W}}_{\mathrm{F}}$is the work done by the force on the suitcase, ${\mathrm{W}}_{\mathrm{g}}$is the work done by the gravity on the suitcase, ${\mathrm{W}}_{\mathrm{N}}$is the work done by the normal force on the suitcase, and ${\mathrm{W}}_{{\mathrm{f}}_{\mathrm{x}}}$is the work done on the suitcase by the friction.

Substitutethe above-obtained values in equation (2) and we get,

$$\begin{gathered} {\mathrm{W}}_{\mathrm{net}}={\mathrm{W}}_{\mathrm{F}}+{\mathrm{W}}_{\mathrm{g}}+{\mathrm{W}}_{\mathrm{N}}+{\mathrm{W}}_{{\mathrm{f}}_{\mathrm{x}}} \\ =608\mathrm{J}+\left(-395\mathrm{J}\right)+0\mathrm{J}+\left(-189\mathrm{J}\right) \\ =24\mathrm{J} \end{gathered}$$

Hence, the net work done on the suitcase is24 J .

If the suitcase has a zero initial speed at the bottom of the ramp and it covers the distance 3.80 m along the ramp then find the speed when it travels the distance 38.0 m along the ramp.

The net force along the vertical direction is,

$$\begin{gathered} \underset{}{鈭�}{\mathrm{F}}_{\mathrm{y}}=\mathrm{N}-\mathrm{尘驳肠辞蝉胃} \\ 0=\mathrm{N}-\mathrm{尘驳肠辞蝉胃} \end{gathered}$$

Thus, the normal force is,

$\mathrm{N}=\mathrm{尘驳肠辞蝉胃}$

The net force along the horizontal direction is,

$$\begin{gathered} \underset{}{鈭�}{\mathrm{F}}_{\mathrm{x}}=\mathrm{F}-\mathrm{尘驳蝉颈苍胃}-{\mathrm{f}}_{\mathrm{k}} \\ \mathrm{ma}=\mathrm{F}-\mathrm{尘驳蝉颈苍胃}-{\mathrm{渭}}_{\mathrm{k}}\mathrm{N} \end{gathered}$$

Substitute $\mathrm{N}=\mathrm{尘驳肠辞蝉胃}$in the above expression we get

$$\begin{gathered} \mathrm{ma}=\mathrm{F}-\mathrm{尘驳蝉颈苍胃}-{\mathrm{渭}}_{\mathrm{k}}\mathrm{尘驳肠辞蝉胃} \\ \mathrm{ma}=\mathrm{F}-\left({\mathrm{渭}}_{\mathrm{k}}\mathrm{尘驳肠辞蝉胃}+\mathrm{尘驳蝉颈苍胃}\right) \end{gathered}$$

Rearrange the above expression for a , such that

$\mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}}-\left({\mathrm{渭}}_{\mathrm{k}}\mathrm{g}\mathrm{肠辞蝉胃}+\mathrm{g}\mathrm{蝉颈苍胃}\right)$鈥︹赌︹赌︹赌︹赌︹赌�.(3)

Substitutethe above-obtained values in equation (3) and we get,

$$\begin{gathered} \mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}}-\left({\mathrm{渭}}_{\mathrm{k}}\mathrm{g}\mathrm{肠辞蝉胃}+\mathrm{g}\mathrm{蝉颈苍胃}\right) \\ =\frac{160\mathrm{N}}{20\mathrm{kg}}-\left(\left(0.3\right)\left(9.8\mathrm{m}路{\mathrm{s}}^{-2}\right)\cos 32掳+\left(9.8\mathrm{m}路{\mathrm{s}}^{-2}\right)\sin 32掳\right) \\ =0.305\mathrm{m}路{\mathrm{s}}^{-2} \end{gathered}$$

Use the below kinematic equation to determine the velocity.

${\mathrm{v}}^2={\mathrm{u}}^2+2\mathrm{as}$鈥︹赌︹赌︹赌︹赌︹赌�(4)

Here, is the final velocity of the suitcase, u is the initial velocity of the suitcase, is the acceleration of the suitcase, and s is the distance traveled by the suitcase.

Substitute 0 m/s for u , $0.305\mathrm{m}路{\mathrm{s}}^{-2}$for a , and 3.80 m for s in equation (4), and we get,

$$\begin{gathered} {\mathrm{v}}^2={\mathrm{u}}^2+2\mathrm{as} \\ {\mathrm{v}}^2=\left(0\mathrm{m}/\mathrm{s}\right)+2\left(0.305\mathrm{m}路{\mathrm{s}}^{-2}\right)\left(3.8\mathrm{m}\right) \\ \mathrm{v}=\sqrt{2.318}\mathrm{m}/\mathrm{s} \\ \mathrm{v}=1.522\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the speed when the suitcase slides 3.80 m along the ramp is $1.522\mathrm{m}/\mathrm{s}$