Q2E - University Physics with Modern Physics

Question

Using a cable with a tension of 1350 N, a tow truck pulls a car 5.00 Km along a horizontal roadway. (a) How much work does the cable do on the car if it pulls horizontally? If it pulls at $\mathbf{35}\mathbf{°}$  above the horizontal? (b) How much work does the cable do on the two trucks in both cases of part (a)? (c) How much work does gravity do on the car in part (a)? 

Step-by-Step Solution

Answer

(a)  6,750,000J (horizontally) and 5,529,276.30 J ( $35°$with the horizontal).

(b)  $-$6,750,000 J (horizontally) and $-$5,529,276.30 J ( $35°$ with the horizontal).

(c) Zero (horizontally and $35°$ ).

Step 1: Identification of the given data

The given data is listed below as-

• The distance covered by the car along a horizontal roadwayis, $\mathrm{d}=5 \mathrm{km}=5,000 \mathrm{m}$  
• The tensionis, $\mathrm{T}=1350 \mathrm{N}$ 

Consider the tow truck and car as point objects.

Step 2: Significance of the work done

Work done on a particle by a constant force i.e., Tension $\overrightarrow{\mathbf{T}}$ during a linear displacement $\overrightarrow{\mathbf{s}}$ is given by

$$\begin{gathered} \mathbf{W}\mathbf{=}\overrightarrow{\mathbf{T}}\mathbf{·}\overrightarrow{\mathbf{d}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{Td}\mathbf{ }\mathbf{cos\varphi } \end{gathered}$$

 

If T and d are in the same direction then $\mathrm{\varphi }=0$ and if it is in opposite direction then $\mathrm{\varphi }=180°$ .

Step 3: Determination of work done by the cable on the car

(a)

When the cable pulls horizontally, the work done by the cable is 

Work done on the car by cable Tension $\mathrm{T}=1350 \mathrm{N}$ will be

${\mathrm{W}}_{\mathrm{car}}=\mathrm{Td} \mathrm{cos\varphi }$ 

Since Force and displacement both are in the same direction, $\mathrm{\varphi }=0$

Here, Tisthe tension in the cable,and d is the displacement covered by the car.

For $\mathrm{T}=1350 \mathrm{N}, \mathrm{d}=5,000 \mathrm{m}$ 

Therefore,

$$\begin{gathered} {\mathrm{W}}_{\mathrm{car}}=\mathrm{Td} \cos 0° \\ =\left(1350\right)\left(5000\right)\left(1\right) \\ =6,750,000 \mathrm{J} \end{gathered}$$

Now, if T makes  $35°$ with the horizontal 

$$\begin{gathered} {\mathrm{W}}_{\mathrm{car}}=\mathrm{Td} \cos 35° \\ =\left(1350\right)\left(5000\right)\left(\cos 35°\right) \\ =5,529,276.30 \mathrm{J} \end{gathered}$$ 

Thus, when the cable pulls horizontally, the work done by the cable on the car is 6,750,000 J   (horizontally) and 5,529,276.30 J ( $35°$ with the horizontal).

(b)

Work done by the cable on the two trucks will be 

$$\begin{gathered} \mathrm{\varphi }=35° \\ \mathrm{\alpha }=\left(180°-35°\right)=145° \\ {\mathrm{W}}_{\mathrm{truck}}=\mathrm{Td} \mathrm{cos\varphi } \end{gathered}$$ 

Since T and d both are in opposite directions, $\mathrm{\varphi }=180°$

$$\begin{gathered} {\mathrm{W}}_{\mathrm{truck}}=\mathrm{Td} \cos 180° \\ =\left(1350\right)\left(5000\right)\left(-1\right) \\ =-6,750,000 \mathrm{J} \end{gathered}$$ 

 

Now, if T makes $35°$ with the horizontal

 

 $$\begin{gathered} {\mathrm{W}}_{\mathrm{truck}}=\mathrm{Td} \cos \left(180°-35°\right) \\ =\left(1350\right)\left(5000\right)\left(\cos 45°\right) \\ =-5,529,276.30 \mathrm{J} \end{gathered}$$

The work is negative since force and displacement are in opposite directions.

Thus, the work done by the cable on the two trucks is $-$6,750,000 J  (horizontally) and $-$5,529,276.30 J (  with the horizontal).

(c)

Work done by gravity on the car will be

The Work done by gravity on the car is zero since the weight of the car is perpendicular to the distance.

Thus, Work done by gravity on the car is Zero (horizontally and $35°$ ).