Problem 48 - College Physics

Question

A person opens a door by applying a 15-N force perpendicular to it at a distance \(0.90 \mathrm{~m}\) from the hinges. The door is pushed wide open (to \(120^{\circ}\) ) in \(2.0 \mathrm{~s}\). (a) How much work was done? (b) What was the average power delivered?

Step-by-Step Solution

Answer

(a) 28.3 J; (b) 14.15 W.

Step 1: Calculate Work Done

Work done is calculated as the torque multiplied by the angle (in radians) through which the force acts. First, calculate the torque (\(\tau\)) using the formula \(\tau = F \times r\), where \(F\) is the force applied (15 N) and \(r\) is the distance from the hinges (0.90 m). So, \(\tau = 15 \times 0.90 = 13.5 \, \text{Nm}\). Next, convert the angle from degrees to radians: \(120^{\circ} = \frac{120 \times \pi}{180} = \frac{2\pi}{3} \, \text{radians}\). Work done \(W\) is then \(\tau \times \theta = 13.5 \times \frac{2\pi}{3} = 28.3 \, \text{J}\).

Step 2: Calculate Average Power

Power is the rate of doing work. It can be calculated using the formula \(P = \frac{W}{t}\), where \(W\) is the work done (28.3 J) and \(t\) is the time taken to do the work (2.0 s). Thus, the average power \(P\) is \(\frac{28.3}{2.0} = 14.15 \, \text{W}\).

Key Concepts

Understanding Torque

Torque is a fundamental concept when studying rotation and mechanics. Imagine torque as the twisting or turning power applied to an object. It's what makes a door swing open when you push on it. To calculate torque (\( \tau \)), you use the formula \( \tau = F \times r \), where:

• \( F \) is the force applied (in Newtons, N).
• \( r \) is the distance from the pivot point or hinge (in meters, m).

For example, if you apply a force of 15 Newtons at a point 0.90 meters away from the hinges of a door, your torque will be \( 15 \times 0.90 = 13.5 \, \text{Nm} \). The longer the distance \( r \), the larger the torque for the same force.
This concept is essential in understanding how rotational forces interact with objects, influencing everything from the nuts and bolts in machinery to the simple act of opening a door.

Calculating Average Power

Average power reflects how fast work is performed over time. In physics, power is the amount of work done per unit of time. The formula to calculate average power \( P \) is \( P = \frac{W}{t} \), where:

• \( W \) represents work done (in Joules, J).
• \( t \) is time (in seconds, s) over which the work is done.

In our door opening example, after calculating that the work done is 28.3 Joules and it took 2 seconds to open the door, the average power is calculated as \( \frac{28.3}{2.0} = 14.15 \, \text{W} \).
Understanding average power allows us to grasp the efficiency and rate at which energy is transferred or converted in a system. This concept is crucial not just in mechanical systems but also in electrical and thermal systems where energy efficiency is key.

The Importance of Conversion of Units

Unit conversion is vital in physics to ensure that calculations are consistent and correct. Many times, the units provided need converting to standard units used in equations. A common example is converting angles from degrees to radians since radians are the standard in most physics calculations involving rotation.
For our door example, the angle is given in degrees (\(120^{\circ}\)). To be accurate in our work calculation, we convert this into radians: \(120^{\circ} = \frac{120 \times \pi}{180} = \frac{2\pi}{3} \, \text{radians}\).
Unit conversion, such as this, maintains the accuracy and reliability of your calculations. It enables seamless understanding and problem-solving across different measurement systems, allowing those involved in science and engineering to communicate and calculate effectively.