Q 11E - University Physics with Modern Physics

Question

Question: A 128.0 N carton is pulled up a frictionless baggage ramp inclined at 30.0^o above the horizontal by a rope exerting a 72.0 N pull parallel to the ramp’s surface. If the carton travels 5.20 m along the surface of the ramp, calculate the work done on it by (a) the rope, (b) by gravity, and (c) the normal force of the ramp. (d) What is the net work done on the carton?

Step-by-Step Solution

Answer

(a)  $W_{rope}=374.4 J$

(b)  $W_g=-332.8 J$

(c)  $W_n=0 J$

(d)  $W_{net}=41.6 J$

Step 1: Identification of the given data

The given data is listed below as-

• The distance covered by cartons along the surface of the ramp is, $s=5.20 m$
• The mass of the baggage is, $w=mg=128 N$
• The coefficient of kinetic friction between the carton and the baggage ramp is ${\mu }_k=0$
• Force due to rope is $F=72 N$
• The inclined angle between the baggage ramp and floor $\varphi =30°$

Step 2: Significance of the work done

Work done can be calculated using the forces exerted on the body and the total displacement of the body. Therefore if a constant force $\overrightarrow{\mathrm{F}}$ is applied during displacement $\overrightarrow{\mathrm{s}}$ along a straight line is given by,

$\begin{array}{rcl}\mathrm{W}& =& \overrightarrow{\mathrm{F}}·\overrightarrow{\mathrm{s}}\\ & =& \mathrm{Fs} \cos \varphi \end{array}$

If F and s are in the same direction then $\varphi =0°$ and if it is in opposite direction then $\varphi =180°$.

Step 3: Determination of work done on the carton by the ramp (a)

The free-body diagram for the carton is shown below:

Take the x-axis along the inclined baggage ramp y-axis perpendicular to the x-axis.

Along the x-axis, two forces are acting on the carton:

One is $F_{rope}$ and another is the component of gravitational force along the x-axis, which is as below

  $F_1=w \cos \theta =mg \cos \theta$

The above two forces are opposite to each other.

Similarly along the y-axis also, two forces are acting on the carton.

One is normal force and another is a component of gravitational force along the y-axis which is as below:

  $F_2=w \sin \theta =mg \sin \theta$

Now, the work done on the carton by the rope will be

$w=Fs \cos \varphi ........\left(1\right)$

Therefore, $W_{rope}=F_{rope}s\cos \varphi$

Here, $F_{rope}$ is the force due to rope, s is the distance covered by carton along the surface of the ramp, and $\varphi$ is the angle between $F_{rope}$ and s.

 For $F_{rope}=72 N$, s = 0, and $\varphi =0°$

 $\begin{array}{rcl}{W}_{rope}& =& 72 N\times 5.2 m \cos 0°\\ & =& 374.4 N·m\\ & =& 374.4 J\end{array}$

Thus, the magnitude of work done on the carton by the ramp is 

Step 4: Determination of work done on the carton due to gravity (b)

The work done by gravity on the carton is given by

 $\begin{array}{rcl}{w}_g& =& F_{g}s \cos \varphi \\ & =& mg s \cos \varphi \end{array}$

The angle between $F_g$ and s is $\varphi =120°$

 $\begin{array}{rcl}{w}_g& =& mg s \cos \varphi \\ & =& 128 N\times 5.2 m\times \cos 120°\\ & =& 128 N\times 5.2 m\times \left(-0.5\right)\\ & =& -332.8 N·m\end{array}$

Thus, the work done by gravity on the carton is given by $-332.8 J$.

Step 5: Determination of work done on the carton due to normal force of the ramp (c)

The work done by normal force on the carton is given by

 $w_n=F_{n}s \cos \varphi$

The angle between $F_n$ and s is 

 $\begin{array}{rcl}{w}_n& =& F_{n}s \cos \varphi \\ w_n& =& F_n\times 5.2\times \cos 90°\\ & =& 0 J\end{array}$

Thus, work done on the carton by normal force is 0J.

Step 6: Determination of net work done on the carton (d)

Net work done on the carton will be the sum of all the individual work done and is given by

 $\begin{array}{rcl}{W}_{net}& =& W_{rope}+W_g+W_n\\ & =& 374.4 J-332.8 J+0 J\\ & =& 41.6 J\end{array}$

  

Thus, the total work done on the carton is 41.6 J.