Q63P - University Physics with Modern Physics

Question

A luggage handler pulls a 20.0-kg suitcase up a ramp inclined at 32.0° above the horizontal by a force $\overrightarrow{\mathbf{F}}$ of magnitude 160 N that acts parallel to the ramp. The coefficient of kinetic friction between the ramp and the incline is ${\mathbf{\mu }}_{\mathbf{k}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{300}$. If the suitcase travels 3.80 m along the ramp, calculate (a) the work done on the suitcase by $\overrightarrow{\mathbf{F}}$ ; (b) the work done on the suitcase by the gravitational force; (c) the work done on the suitcase by the normal force; (d) the work done on the suitcase by the friction force; (e) the total work done on the suitcase. (f) If the speed of the suitcase is zero at the bottom of the ramp, what is its speed after it has traveled 3.80 m along the ramp?

Step-by-Step Solution

Answer

1. The work done on the suitcase by the force is 680 J .
2. The work on the suitcase by gravity is  $-395 \mathrm{J}$.
3. The work done on the suitcase by the normal force is 0 J .
4. The work done on the suitcase by the friction is $-189 \mathrm{J}$ .
5. The net work done on the suitcase is 24 J .
6. The speed when the suitcase slides 3.80 m along the ramp is 1.522 m/s

Step 1: Given data

Mass of the suitcase is m = 20.0 kg

Inclined of the ramp is $\theta =32°$  

The magnitude of the force is F = 160 N 

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

The distance along the ramp is s = 3.80 m 

Step 2: Solution

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,

$$\begin{gathered} \mathbf{W}\mathbf{=}\overrightarrow{\mathbf{F}}\mathbf{·}\overrightarrow{\mathbf{d}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{Fdcos\varphi } \end{gathered}$$ ……………….(1)

Here, $\overrightarrow{\mathbf{F}}$ is the force exerted on the object, $\overrightarrow{\mathbf{d}}$ is the distance covered by the object, and $\mathbf{\theta }$ is the angle between force and displacement.

Step 3: (a) Find the work done on the suitcase by

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 $\mathrm{\varphi }=0°$ .

 

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

 $$\begin{gathered} {\mathrm{W}}_{\mathrm{F}}=\mathrm{Fscos\varphi } \\ =\left(160 \mathrm{N}\right)\left(3.8 \mathrm{m}\right)\left(\cos 0°\right) \\ =680 \mathrm{J} \end{gathered}$$

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

Step 4: (b) Find the work done on the suitcase by the gravitational force

Force due to gravity,

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

$$\begin{gathered} \mathrm{\varphi }=90°+32° \\ =122° \end{gathered}$$ 

 

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

${\mathrm{W}}_{\mathrm{g}}=\left(\mathrm{mg}\right)\mathrm{scos\varphi }$ 

 

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

 

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

 

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

Step 5: (c) Find the work done on the suitcase by the normal force

Since the normal force ${\mathrm{F}}_{\mathrm{N}}$ is equal to $\mathrm{mgcos\theta }$ .

 

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

${\mathrm{W}}_{\mathrm{N}}=\left(\mathrm{mgcos\theta }\right)\mathrm{scos\varphi }$ 

 

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

 

$$\begin{gathered} {\mathrm{W}}_{\mathrm{N}}=\left(\mathrm{mgcos\theta }\right)\mathrm{scos\varphi } \\ =\left(20 \mathrm{kg}\right)\left(9.8 \mathrm{m}·{\mathrm{s}}^{-2}\right)\left(\cos 32°\right)\left(3.8 \mathrm{m}\right)\left(\cos 90°\right) \\ =0 \mathrm{J} \end{gathered}$$ 

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

Step 6: (d) Find the work done on the suitcase by the friction force

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

 

${\mathrm{W}}_{{\mathrm{f}}_{\mathrm{k}}}={\mathrm{f}}_{\mathrm{k}}\mathrm{scos\varphi }$ 

 

Since kinetic frictional force ${\mathrm{f}}_{\mathrm{k}}$ is equal to ${\mathrm{\mu }}_{\mathrm{k}}\mathrm{mgcos\theta }$ . So, the work done is,

${\mathrm{W}}_{{\mathrm{f}}_{\mathrm{x}}}=\left({\mathrm{\mu }}_{\mathrm{k}} \mathrm{mgcos\theta }\right)\mathrm{scos\varphi }$ 

 

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

  $$\begin{gathered} {\mathrm{W}}_{{\mathrm{f}}_{\mathrm{x}}}=\left({\mathrm{\mu }}_{\mathrm{k}} \mathrm{mgcos\theta }\right)\mathrm{scos\varphi } \\ =\left(0.3\right)\left(20 \mathrm{kg}\right)\left(9.8 \mathrm{m}·{\mathrm{s}}^{-2}\right)\left(\cos 32°\right)\left(3.8 \mathrm{m}\right)\left(\cos 180°\right) \\ =-189 \mathrm{J} \end{gathered}$$

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

Step 7: (e) Find the total work done on the suitcase

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.

 

Substitute the 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 is 24 J .

Step 8: (f) Find what is its speed after it has traveled 3.80 m along the ramp

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, 

 

$$\begin{gathered} \sum _{ }{\mathrm{F}}_{\mathrm{y}}=\mathrm{N}-\mathrm{mgcos\theta } \\ 0=\mathrm{N}-\mathrm{mgcos\theta } \end{gathered}$$ 

 

Thus, the normal force is,

 

$\mathrm{N}=\mathrm{mgcos\theta }$ 

 

The net force along the horizontal direction is,

 

$$\begin{gathered} \sum _{ }{\mathrm{F}}_{\mathrm{x}}=\mathrm{F}-\mathrm{mgsin\theta }-{\mathrm{f}}_{\mathrm{k}} \\ \mathrm{ma}=\mathrm{F}-\mathrm{mgsin\theta }-{\mathrm{\mu }}_{\mathrm{k}}\mathrm{N} \end{gathered}$$ 

 

Substitute $\mathrm{N}=\mathrm{mgcos\theta }$ in the above expression we get 

 

$$\begin{gathered} \mathrm{ma}=\mathrm{F}-\mathrm{mgsin\theta }-{\mathrm{\mu }}_{\mathrm{k}}\mathrm{mgcos\theta } \\ \mathrm{ma}=\mathrm{F}-\left({\mathrm{\mu }}_{\mathrm{k}}\mathrm{mgcos\theta }+\mathrm{mgsin\theta }\right) \end{gathered}$$ 

 

Rearrange the above expression for a , such that 

 

$\mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}}-\left({\mathrm{\mu }}_{\mathrm{k}}\mathrm{g} \mathrm{cos\theta }+\mathrm{g} \mathrm{sin\theta }\right)$ ……………….(3)

 

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

 

$$\begin{gathered} \mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}}-\left({\mathrm{\mu }}_{\mathrm{k}}\mathrm{g} \mathrm{cos\theta }+\mathrm{g} \mathrm{sin\theta }\right) \\ =\frac{160 \mathrm{N}}{20 \mathrm{kg}}-\left(\left(0.3\right)\left(9.8 \mathrm{m}·{\mathrm{s}}^{-2}\right)\cos 32°+\left(9.8 \mathrm{m}·{\mathrm{s}}^{-2}\right)\sin 32°\right) \\ =0.305 \mathrm{m}·{\mathrm{s}}^{-2} \end{gathered}$$ 

 

Use the below kinematic equation to determine the velocity.

 

${\mathrm{v}}^2={\mathrm{u}}^2+2\mathrm{as}$ ………………(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 , $0.305 \mathrm{m}·{\mathrm{s}}^{-2}$ for  a , and 3.80 m for s in equation (4), and we get,

 

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

 

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