Question

A tow truck pulls a car by a cable that makes an angle of 30° to the horizontal. The force applied to the tow cable is 7.2×10^3 N. How much work is done in pulling the car horizontally 5.5 km?

Answer 1

Given,

The angle made by the cable, θ=30°

The force applied to the cable, F=7.2×10³ N

The distance to which the car was pulled, d=5.5 km=5.5×10³ m

The force with which the car was pulled by the cable is equal to the horizontal component of the force applied by the tow truck, i.e., the force applied to the cable.

Thus the force with which the car was pulled is given by,

`f=F\cos \theta`

On applying the known values,

`\begin{gathered} f=7.2*10^3*\cos 30^(\circ) \\ =6.24*10^3\text{ N} \end{gathered}`

The work done in pulling the car is equal to the product of the force with which the car was pulled and the distance to which the car was pulled.

That is,

`W=fd`

On substituting the known values,

`\begin{gathered} w=6.24*10^3*5.5*10^3 \\ =34.32*10^6\text{ J} \end{gathered}`

Thus the work done in pulling the car horizontally is 34.32×10⁶ J