Mechanics 1: Power
Welcome to the study notes on Power! If you have ever wondered why a sports car accelerates faster than a family hatchback, or why a cyclist has to pedal much harder to maintain their speed up a steep hill, you are already thinking about power. In this chapter, we will connect what you already know about forces, work, and motion to understand how engines and moving objects generate and use power. Don't worry if mechanics sometimes feels heavy—we will break everything down step-by-step!
1. What is Power?
In everyday conversation, "power" can mean strength or authority. In mechanics, however, power has a very precise mathematical definition: it is the rate at which work is done, or the rate at which energy is transferred.
If an amount of work \(W\) is done in a time \(t\), the average power \(P\) is given by:
\(P = \frac{W}{t}\)
Units of Power:
• Work is measured in Joules (\(\text{J}\)).
• Time is measured in seconds (\(\text{s}\)).
• Therefore, power is measured in Joules per second (\(\text{J s}^{-1}\)), which is given the special name Watts (\(\text{W}\)).
• In many exam questions, power is given in kilowatts (\(\text{kW}\)). Always remember to convert to Watts before calculating: \(1\text{ kW} = 1000\text{ W}\).
Did you know? The unit "Watt" is named after the Scottish engineer James Watt, who also coined the term "horsepower" to compare the output of steam engines with the draft horses of his day!
Key Takeaway: Power tells you how fast work is being delivered. Doing \(1000\text{ J}\) of work in \(1\text{ second}\) requires ten times as much power as doing the same work in \(10\text{ seconds}\).
2. The Core Formula: \(P = Fv\)
When a vehicle moves at an instantaneous velocity \(v\) under a driving force (also called the tractive force) \(F\), we can find the instantaneous power produced by the engine.
Recall that work done is given by force multiplied by distance: \(W = F \times s\).
Substituting this into our power definition gives:
\(P = \frac{W}{t} = \frac{F \times s}{t} = F \times \left(\frac{s}{t}\right)\)
Since speed is distance over time (\(v = \frac{s}{t}\)), we obtain the fundamental relationship for this chapter:
\(P = Fv\)
Where:
• \(P\) is the power produced by the driving force (in \(\text{W}\)).
• \(F\) is the tractive force (driving force) exerted by the engine (in \(\text{N}\)).
• \(v\) is the instantaneous speed of the vehicle (in \(\text{m s}^{-1}\)).
A Very Important Distinction: Tractive Force vs. Resultant Force
Warning: One of the most common mistakes students make is confusing the tractive force \(F\) with the resultant (net) force \(F_{\text{net}}\).
• \(F\) in the formula \(P = Fv\) is only the forward force generated by the engine.
• \(F_{\text{net}} = ma\) is the sum of all forces acting on the body (engine driving force minus resistances).
Rearranging the power formula gives the tractive force at any speed:
\(F = \frac{P}{v}\)
Notice that for a fixed power output, as speed \(v\) increases, the available driving force \(F\) decreases. This is why a car accelerates quickly at low speeds, but its acceleration drops off as it goes faster!
Key Takeaway: At any speed \(v\), an engine working at power \(P\) delivers a forward driving force of \(F = \frac{P}{v}\).
3. Motion on a Horizontal Road
Let us look at a vehicle of mass \(m\) moving along a flat, horizontal road with an engine operating at power \(P\). The vehicle experiences a forward driving force \(F = \frac{P}{v}\) and a total resistance force \(R\) (such as air resistance and friction).
Applying Newton's Second Law (\(F_{\text{net}} = ma\)) in the direction of motion:
\(F - R = ma\)
\(\frac{P}{v} - R = ma\)
Case 1: Finding Instantaneous Acceleration
When you know the power \(P\), speed \(v\), mass \(m\), and resistance \(R\), you can calculate the acceleration at that exact moment:
\(a = \frac{\frac{P}{v} - R}{m}\)
Case 2: Maximum (Terminal) Speed
A vehicle reaches its maximum speed when it can no longer accelerate, meaning \(a = 0\). When acceleration is zero, the driving force exactly balances the total resistance:
\(F = R \implies \frac{P}{v_{\text{max}}} = R \implies P = R v_{\text{max}}\)
Worked Example 1: Horizontal Motion
Problem: A car of mass \(1200\text{ kg}\) travels along a straight horizontal road. The engine operates at a constant power of \(48\text{ kW}\), and the resistance to motion is a constant \(600\text{ N}\).
(a) Find the acceleration of the car when its speed is \(20\text{ m s}^{-1}\).
(b) Find the maximum speed of the car.
Solution:
Step 1: Convert power to standard SI units.
\(P = 48\text{ kW} = 48000\text{ W}\).
Part (a): Acceleration at \(v = 20\text{ m s}^{-1}\)
Find the tractive force at this speed:
\(F = \frac{P}{v} = \frac{48000}{20} = 2400\text{ N}\)
Apply Newton's Second Law in the forward direction:
\(F - R = ma\)
\(2400 - 600 = 1200a\)
\(1800 = 1200a\)
\(a = \frac{1800}{1200} = 1.5\text{ m s}^{-2}\).
Part (b): Maximum Speed
At maximum speed, acceleration \(a = 0\), so tractive force equals resistance:
\(F = R = 600\text{ N}\)
Using \(v_{\text{max}} = \frac{P}{F}\):
\(v_{\text{max}} = \frac{48000}{600} = 80\text{ m s}^{-1}\).
Key Takeaway: On a horizontal surface, maximum speed occurs when the tractive force equals the resistance force: \(\frac{P}{v} = R\).
4. Motion on an Inclined Plane (Slopes)
When a vehicle moves up or down a slope inclined at an angle \(\theta\) to the horizontal, gravity enters the picture. We must resolve the weight \(mg\) parallel to the plane.
The component of weight acting down the slope is always \(mg \sin \theta\).
Vehicle Moving UP a Slope
When moving uphill, both the resistance \(R\) and the component of weight \(mg \sin \theta\) pull back against the vehicle.
Equation of motion up the slope:
\(F - R - mg \sin \theta = ma\)
\(\frac{P}{v} - R - mg \sin \theta = ma\)
At maximum speed up the slope (\(a = 0\)):
\(\frac{P}{v_{\text{max}}} = R + mg \sin \theta\)
Vehicle Moving DOWN a Slope
When moving downhill, gravity helps the vehicle! The component of weight acts in the direction of motion.
Equation of motion down the slope:
\(F + mg \sin \theta - R = ma\)
\(\frac{P}{v} + mg \sin \theta - R = ma\)
At maximum speed down the slope (\(a = 0\)):
\(\frac{P}{v_{\text{max}}} + mg \sin \theta = R \implies \frac{P}{v_{\text{max}}} = R - mg \sin \theta\)
Note: If a vehicle is "coasting" (freewheeling) with the engine turned off, then \(P = 0\) and \(F = 0\).
Worked Example 2: Motion on a Slope
Problem: A van of mass \(1000\text{ kg}\) is driven up a hill inclined at an angle \(\theta\) to the horizontal, where \(\sin \theta = \frac{1}{14}\). The engine works at a constant rate of \(35\text{ kW}\) and experiences a constant non-gravitational resistance of \(500\text{ N}\). Take \(g = 9.8\text{ m s}^{-2}\).
(a) Calculate the maximum steady speed at which the van can travel up the hill.
(b) The van reaches the crest and now travels down the same slope with the engine still working at \(35\text{ kW}\) and the same resistance. Find its acceleration when its speed is \(25\text{ m s}^{-1}\).
Solution:
Step 1: Identify the constants and components.
\(m = 1000\text{ kg}\)
\(P = 35000\text{ W}\)
\(R = 500\text{ N}\)
Weight component down the slope: \(mg \sin \theta = 1000 \times 9.8 \times \frac{1}{14} = 700\text{ N}\).
Part (a): Maximum speed uphill
At steady maximum speed, \(a = 0\). Resolving up the slope:
\(F_{\text{up}} = R + mg \sin \theta\)
\(F_{\text{up}} = 500 + 700 = 1200\text{ N}\)
Now find speed using \(v = \frac{P}{F}\):
\(v_{\text{max}} = \frac{35000}{1200} = \frac{175}{6} \approx 29.2\text{ m s}^{-1}\) (to \(3\) significant figures).
Part (b): Acceleration downhill at \(v = 25\text{ m s}^{-1}\)
First, find the tractive force generated by the engine at this speed:
\(F_{\text{down}} = \frac{P}{v} = \frac{35000}{25} = 1400\text{ N}\)
Apply Newton's Second Law down the slope:
\(F_{\text{down}} + mg \sin \theta - R = ma\)
\(1400 + 700 - 500 = 1000a\)
\(1600 = 1000a\)
\(a = \frac{1600}{1000} = 1.6\text{ m s}^{-2}\).
Key Takeaway: Always draw a clear diagram showing all forces parallel to the slope. Be careful with signs: weight opposes uphill motion (\(- mg \sin \theta\)) and assists downhill motion (\(+ mg \sin \theta\)).
5. Common Pitfalls to Avoid
1. Unit Neglect: Always convert kilowatts to Watts (\(\times 1000\)) and kilometers per hour to meters per second (\(\div 3.6\)) before substituting into equations.
2. Replacing \(F\) with \(ma\) in \(P = Fv\): Never write \(P = (ma)v\). The force \(F\) in \(P = Fv\) is the driving force alone, not the resultant force. The correct process is: find \(F = \frac{P}{v}\) first, then set up the equation of motion \(F - \text{Resistances} = ma\).
3. Forgetting the Slope Component: When a vehicle travels on an incline, the engine must overcome both the frictional/air resistance and the gravity component \(mg \sin \theta\).
4. Confusing Sine and Cosine: The component of weight parallel to a slope of angle \(\theta\) to the horizontal is always \(mg \sin \theta\). The component perpendicular to the slope is \(mg \cos \theta\).
6. Quick Chapter Summary
• Definition of Power: \(P = \frac{W}{t}\) (measured in Watts, where \(1\text{ W} = 1\text{ J s}^{-1}\)).
• Tractive Force & Speed Formula: \(P = Fv \iff F = \frac{P}{v}\).
• Horizontal Motion Equation: \(\frac{P}{v} - R = ma\).
• Motion Up a Slope: \(\frac{P}{v} - R - mg \sin \theta = ma\).
• Motion Down a Slope: \(\frac{P}{v} + mg \sin \theta - R = ma\).
• Maximum / Terminal Speed: Set acceleration \(a = 0\) and solve for \(v\).