Chapter 10, Problem 63A

To determine

Power developed by conveyor belt.

Answer to Problem 63A

  $P=367.5\text{W}$

Explanation of Solution

Given:

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

Inclination angle is $\theta =30°$ .

Mass of newspaper is $m=1\text{kg}$ .

Number of bundles is 15.

Each bundle has 25 newspapers.

Time is $t=1\text{min}=60\text{s}$ .

Formula Used:

Power developed can be obtained by:

  $P=\frac{W}{t}=\frac{Fd}{t}$

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

Calculation:

  

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

As the newspapers are 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,

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

Now, number of newspapers are $25\times 15=375$

So, mas of newspaper is

  $\begin{array}{c}m=375\times 1\text{kg}\\ =375\text{kg}\end{array}$

On putting the values in (1)

  $\begin{array}{c}P=\frac{\left(375\times 9.8\times \sin 30°\right)\times 12}{60}\\ =367.5\text{W}\end{array}$

Conclusion:

Therefore,power developed by conveyor belt is $P=367.5\text{W}$ .