Chapter 10, Problem 50A

To determine

Power developed by tractor.

Answer to Problem 50A

  $2.023\text{kW}$

Explanation of Solution

Given:

Mass of tractor is $m=120\text{kg}$ .

Length of hill is $d=12\text{m}$ .

Angle of inclination of hill is $\theta ={21}^{\text{o}}$ .

Time taken is $t=2.5\text{s}$ .

Formula used:

Power developed can be obtained by:

  $P=\frac{W}{t}=\frac{Fd}{t}$

Where, $P$ is power developed, $F$ is the applied force, $d$ is displacement in the direction of force and $t$ is time taken to do work.

Calculation:

The diagram from the given condition is,

  

Here, component of weight $mg$ along the ramp is $mg\sin \theta$ as shown.

As the tractor is moved with constant velocity, it means there is no net force parallel to ramp.

So, the force parallel to ramp is $F=mg\sin \theta$

So,

  $P=\frac{\left(mg\sin \theta \right)d}{t}$

On putting the values,

  $\begin{array}{c}P=\frac{\left(120\times 9.8\times \sin 21°\right)\times 12}{2.5}\\ =2023\text{W}\\ =2.023\times {10}^3\text{W}\end{array}$

Conclusion:

Therefore,power developed by tractor is $P=2.023\text{kW}$ .