Q66P - University Physics with Modern Physics

Question

A 5.00-kg package slides 2.80 m down a long ramp that is inclined at 24.0° below the horizontal. The coefficient of kinetic friction between the package and the ramp is ${\mathit{\mu }}_{\mathbf{k}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{310}$ . Calculate (a) the work done on the package by friction; (b) the work done on the package by gravity; (c) the work done on the package by the normal force; (d) the total work done on the package. (e) If the package has a speed of 2.20 m/s at the top of the ramp, what is its speed after it has slid 2.80 m down the ramp?

Step-by-Step Solution

Answer

1.  The work done on the package by the frictional force is $-38.85 \mathrm{J}$ .
2.  The work done on the package by gravity is 55.8 J .
3.  The work done on the package by the normal force is 0 J .
4.  The net work done on the package is 16.95 J .
5.  The speed when the block slid 2.8 m down the ramp is 3.41 m/s .

Step 1: Concept Introduction

The work done by any object is given by the multiplication of force and displacement vectors, such that,

 

work done$\mathbf{=}\overrightarrow{\mathbf{F}}\mathbf{.}\overrightarrow{\mathbf{d}}$

Step 2: Given data

A 5.00-kg package slides 2.80 m down

 

Coefficient of friction ${\mathrm{\mu }}_{\mathrm{k}}=0.310$ 

Step 3: Solution

The free-body diagram for the package is shown as,

In this diagram, a package of mass m slides the distance s down a ramp that is inclined at an angle $\theta$ below the horizontal. So, the angle above the horizontal will also be $\theta$ (alternate interior angle property). The coefficient of kinetic friction between the package and the ramp surface is ${\mathrm{\mu }}_{\mathrm{k}}$ . and ${\mathrm{f}}_{\mathrm{k}}$  is the force due to the friction and g is the acceleration due to gravity.

Step 4: (a) Find the work done on the package by friction

Work done on the package by the friction is given by,

 

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

 

Substitute ${\mathrm{f}}_{\mathrm{k}}={\mathrm{\mu }}_{\mathrm{k}}\mathrm{mgcos\theta }$, in the above expression.

 

${\mathrm{w}}_{{\mathrm{f}}_{\mathrm{k}}}=\left({\mathrm{\mu }}_{\mathrm{k}}\mathrm{mgcos\theta }\right)\mathrm{s} \mathrm{cos\varphi }$ 

 

Substitute 5 kg for m, 2.8m for s, 0.31 for ${\mu }_k, 24°$ for $\theta 180°$ for $\varphi$ and $9.8 m/s^2$ for g in the above expression.

 

$$\begin{gathered} {\mathrm{w}}_{{\mathrm{f}}_{\mathrm{x}}}=\left({\mathrm{\mu }}_{\mathrm{k}}\mathrm{mgcos\theta }\right)\mathrm{s} \mathrm{cos\varphi } \\ =\left(0.31\times 5 \mathrm{kg}\times 9.8 \mathrm{m}/{\mathrm{s}}^2\times \cos 24°\right) \\ =-38.85 \mathrm{J} \end{gathered}$$ 

 

Hence, the work done on the package by the frictional force is $-38.85 \mathrm{J}$.

Step 5: (b) Find the work done on the package by gravity

Work done by the gravity on the package is given by.

 

${\mathrm{w}}_{\mathrm{g}}={\mathrm{F}}_{\mathrm{g}}\mathrm{scos\varphi }$ 

 

Substitute ${\mathrm{F}}_{\mathrm{g}}=\mathrm{mg}$ in the above equation.

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

 

Substitute  5 kg for $m$ , $9.8 \mathrm{m}/{\mathrm{s}}^2$ for g , 2.8 m for s , and $66°$ for $\mathrm{\varphi }$ in the above equation,

 

$$\begin{gathered} {\mathrm{w}}_{\mathrm{g}}=\left(\mathrm{mg}\right)\mathrm{scos\varphi } \\ =\left(5 \mathrm{kg}\times 9.8 \mathrm{m}/{\mathrm{s}}^2\right)\left(2.8 \mathrm{m}\right)\left(\cos 66°\right) \\ =55.8 \mathrm{J} \end{gathered}$$ 

 

Hence, the work done on the package by gravity is $55.8 \mathrm{J}$.

Step 6: (c) Find the work done on the package by the normal force

Work done by the normal force on the package is given by,

 

${\mathrm{w}}_{\mathrm{N}}={\mathrm{F}}_{\mathrm{N}} \mathrm{scos\varphi }$ 

 

Substitute 5 kg for m , 2.8 m for s , $24°$ for $\theta$ , $90°$ for $\mathrm{\varphi }$ , and $9.8 \mathrm{m}/{\mathrm{s}}^2$ for $g$ in the above equation

 

$$\begin{gathered} {\mathrm{w}}_{\mathrm{N}}=\left(\mathrm{mg} \mathrm{cos\theta }\right)\mathrm{scos\varphi } \\ {\mathrm{w}}_{\mathrm{N}}=\left(5 \mathrm{kg}\times 9.8 \mathrm{m}/{\mathrm{s}}^2\times \cos 24°\right)\left(2.8 \mathrm{m}\right)\left(\cos 90°\right) \\ =0 \mathrm{J} \end{gathered}$$ 

 

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

Step 7: (d) Find the total work done on the package

The net work done on the package is

 

${\mathrm{w}}_{\mathrm{net}}={\mathrm{w}}_{\mathrm{F}}+{\mathrm{w}}_{\mathrm{g}}+{\mathrm{w}}_{\mathrm{N}}$ 

 

Here, ${\mathrm{w}}_{\mathrm{F}}$ is the work done by the force on the package, ${\mathrm{w}}_{\mathrm{g}}$ is the work done by the gravity on the package, and ${\mathrm{w}}_{\mathrm{N}}$ is the work done by the normal force on the package.

 

Substitute $-38.85 \mathrm{J}$  for ${\mathrm{w}}_{\mathrm{F}}$ , 55.8 J  for ${\mathrm{w}}_{\mathrm{g}}$ , and 0 J for  ${\mathrm{w}}_{\mathrm{N}}$

$$\begin{gathered} {\mathrm{w}}_{\mathrm{net}}={\mathrm{w}}_{\mathrm{F}}+{\mathrm{w}}_{\mathrm{g}}+{\mathrm{w}}_{\mathrm{N}} \\ =-38.85 \mathrm{J}+55.8 \mathrm{J}+0 \mathrm{J} \\ =16.95 \mathrm{J} \end{gathered}$$ 

 

Hence, the net work done on the package is $16.95 \mathrm{J}$  .

Step 8: (e) Find the speed after it has slid 2.80 m down the ramp

If the package has an initial speed of 2.2 m/s on the top of the ramp and it covers the distance of 2.8 m  down the ramp

 

The free-body diagram of the package for the above problem is given as,

From the free body diagram, 

 

$\mathrm{ma}=\mathrm{mgsin\theta }-{\mathrm{\mu }}_{\mathrm{k}}\mathrm{mgcos\theta }$ 

 

Rewrite the above equation,

 

$\mathrm{a}=\mathrm{g}\left(\mathrm{sin\theta }-{\mathrm{\mu }}_{\mathrm{k}} \mathrm{cos\theta }\right)$ 

 

Substitute 0.31  for ${\mu }_k$ , $9.8m/s^2$ for $g$ , and $24°$ for $\theta$ in the above equation,

 

$$\begin{gathered} \mathrm{a}=\mathrm{g}\left(\mathrm{sin\theta }-{\mathrm{\mu }}_{\mathrm{k}} \mathrm{cos\theta }\right) \\ \mathrm{a}=\left(9.8 \mathrm{m}/{\mathrm{s}}^2\right)\left(\sin 24°-\left(0.31\right)\left(\cos 24°\right)\right) \\ =1.22 \mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$ 

 

 

Use the equation of kinematics,

 

$v^2=u^2+2as$ 

 

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

 

Substitute $2.2 \mathrm{m}/{\mathrm{s}}^2$  for u , $1.22\mathrm{m}/{\mathrm{s}}^2$  for a , and 2.8 m  for s .

 

$$\begin{gathered} v^2={\left(2.2 \mathrm{m}/\mathrm{s}\right)}^2+\left(2\left(1.22 \mathrm{m}/{\mathrm{s}}^2\right)\left(2.8 \mathrm{m}\right)\right) \\ v=\sqrt{4.84+6.78}\mathrm{m}/\mathrm{s} \\ v=\sqrt{11.62} \mathrm{m}/\mathrm{s} \\ v=3.14 \mathrm{m}/\mathrm{s} \end{gathered}$$ 

 

Hence, the speed when the block slid 2.8 m down the ramp is $3.14 \mathrm{m}/\mathrm{s}$ .