Power: Videos & Practice Problems

Power is defined as the rate at which work is done or energy is transferred, expressed mathematically as $P=\frac{W}{\mathrm{\Delta t}}$ or $P=\frac{E}{T}$, where P is power, W is work, E is energy, and T is time. The unit of power is the watt (W), equivalent to joules per second (J/s). Understanding power is crucial for calculating energy consumption, such as in a 100 W light bulb using 360,000 joules in one hour, or determining the average power delivered by a car engine during acceleration.

Power

In the study of physics, understanding the concept of power is essential as it relates to how quickly work is done or energy is transferred. Power is defined as the average power, which can be expressed mathematically as:

\( P_{\text{avg}} = \frac{W}{\Delta t} \)

Here, \( P_{\text{avg}} \) represents average power, \( W \) is the work done, and \( \Delta t \) is the change in time. This equation highlights that power measures the rate at which work is performed or energy is transferred. Since work is a transfer of energy, this relationship can also be represented as:

\( P_{\text{avg}} = \frac{E}{\Delta t} \)

where \( E \) is the energy. The unit of power is the watt (W), which is equivalent to one joule per second (J/s). It is important to distinguish between the symbol for power (W) and the symbol for work (also W), as they represent different concepts.

To illustrate the application of this concept, consider a 100-watt light bulb. To find the energy consumed in one hour, we first convert the time from hours to seconds:

\( 1 \text{ hour} = 3600 \text{ seconds} \)

Using the power equation, we can calculate the energy:

\( E = P \cdot \Delta t = 100 \, \text{W} \times 3600 \, \text{s} = 360,000 \, \text{J} \)

This example demonstrates how to use the power equation to find energy consumption over time.

Next, consider a scenario involving a 1300 kg sports car accelerating from rest to a velocity of 40 m/s over a time interval of 7 seconds. To find the average power delivered by the engine, we can use the work-energy principle, which states that the net work done on an object is equal to the change in its kinetic energy:

\( \Delta KE = \frac{1}{2} m v_{\text{final}}^2 - \frac{1}{2} m v_{\text{initial}}^2 \)

Since the initial velocity is zero, the equation simplifies to:

\( \Delta KE = \frac{1}{2} m v_{\text{final}}^2 \)

Substituting the known values:

\( \Delta KE = \frac{1}{2} \times 1300 \, \text{kg} \times (40 \, \text{m/s})^2 \)

Now, to find the average power, we divide the work done (which is equal to the change in kinetic energy) by the time interval:

\( P_{\text{avg}} = \frac{\Delta KE}{\Delta t} = \frac{\frac{1}{2} \times 1300 \times 1600}{7} \)

Calculating this gives:

\( P_{\text{avg}} \approx 1.49 \times 10^5 \, \text{W} \) or 149 kW.

This example illustrates how to apply the concepts of work and energy to determine the power output of an engine during acceleration. Understanding these relationships is crucial for solving various physics problems related to energy transfer and motion.

How much power must an elevator motor supply in order to lift a 1000 kg elevator at constant speed a height of 100m in 50 seconds?

• A. 100,000 W
• B. 0 W
• C. 2,000 W
• D. 19,600 W

Answer: (D)

Power of Pushing a Box

In this scenario, we analyze a 15-kilogram box sliding across a frictionless surface, initially moving at a speed of 10 meters per second and accelerating to a final speed of 30 meters per second with a constant acceleration of 2 meters per second squared. The primary focus is on calculating the change in mechanical energy, denoted as ΔME, which encompasses the change in kinetic energy since there is no potential energy change due to the absence of gravity and springs.

The formula for kinetic energy (KE) is given by:

KE = \frac{1}{2} mv^2

where m is mass and v is velocity. The change in mechanical energy can be expressed as:

ΔME = KE_{final} - KE_{initial} = \frac{1}{2} m v_{final}^2 - \frac{1}{2} m v_{initial}^2

Substituting the known values:

ΔME = \frac{1}{2} (15 \, \text{kg}) (30 \, \text{m/s})^2 - \frac{1}{2} (15 \, \text{kg}) (10 \, \text{m/s})^2

Calculating this yields:

ΔME = 6,000 \, \text{joules}

This change in mechanical energy indicates that work is done on the box, resulting in an increase in kinetic energy by 6,000 joules.

Next, we determine the average power, which is the rate at which energy is transferred. Power (P) is defined as:

P = \frac{ΔE}{Δt}

Here, ΔE represents the change in mechanical energy, and Δt is the time taken for this change. To find the time, we utilize kinematic equations. Given the initial velocity (v₀ = 10 m/s), final velocity (v_f = 30 m/s), and acceleration (a = 2 m/s²), we can apply the first kinematic equation:

v_f = v₀ + at

Rearranging for time gives:

t = \frac{v_f - v₀}{a} = \frac{30 \, \text{m/s} - 10 \, \text{m/s}}{2 \, \text{m/s}^2} = 10 \, \text{seconds}

Now, substituting the values into the power equation:

P = \frac{6,000 \, \text{joules}}{10 \, \text{seconds}} = 600 \, \text{watts}

Thus, the average power exerted on the box during its acceleration is 600 watts. This analysis illustrates the relationship between work, energy, and power in a system experiencing uniform acceleration.

Power of a Winch on an Incline

In this scenario, we analyze a block with a mass of 1400 kilograms being pulled up an incline by a winch and cable. The goal is to determine the power exerted by the cable, which is referred to as tension. To find the power, we can use the equation for power, which relates energy change over time, or equivalently, work done over time. The work done is calculated as the product of force and distance, specifically:

$$ P = \frac{W}{\Delta t} = \frac{F \cdot d \cdot \cos(\theta)}{\Delta t} $$

In this case, the angle between the force and the distance is zero degrees since both the tension and the distance are directed up the incline, simplifying the cosine term to 1. We know the time interval, Δt, is 25 seconds, but we need to determine both the tension force and the distance the block is pulled up the incline.

Given the incline angle of 37 degrees and the coefficient of kinetic friction (μ_k) of 0.4, we can analyze the forces acting on the block. Since the block is moving at a constant speed, the net acceleration is zero, allowing us to set up the equation:

$$ T - mg \sin(\theta) - F_{friction} = 0 $$

Where:

• $$ F_{friction} = \mu_k \cdot F_{normal} = \mu_k \cdot mg \cos(\theta) $$

Substituting the known values into the equation, we can express the tension (T) as:

$$ T = mg \sin(\theta) + \mu_k \cdot mg \cos(\theta) $$

Plugging in the values:

• Mass (m) = 1400 kg
• Acceleration due to gravity (g) = 9.8 m/s²
• Angle (θ) = 37 degrees

Calculating the components:

• $$ mg \sin(37^\circ) = 1400 \cdot 9.8 \cdot \sin(37^\circ) $$
• $$ F_{friction} = 0.4 \cdot 1400 \cdot 9.8 \cdot \cos(37^\circ) $$

After performing the calculations, we find that the tension in the cable is 12640 newtons.

Next, we need to determine the distance (d) the block is pulled up the incline. Given the height (h) of the incline is 60.2 meters, we can use the sine function to relate the height to the hypotenuse (d):

$$ \sin(\theta) = \frac{h}{d} $$

Rearranging gives us:

$$ d = \frac{h}{\sin(\theta)} $$

Substituting the known values:

$$ d = \frac{60.2}{\sin(37^\circ)} $$

Calculating this yields a distance of 100 meters.

Now that we have both the tension and the distance, we can calculate the power:

$$ P = \frac{T \cdot d}{\Delta t} = \frac{12640 \cdot 100}{25} $$

This results in a power output of 50,560 watts, which can also be expressed as 50.5 kilowatts. This indicates the significant amount of energy required by the motor to pull the block up the incline.

Here’s what students ask on this topic:

What is the definition of power in physics?

Power in physics is defined as the rate at which work is done or energy is transferred. Mathematically, it is expressed as $P=\frac{W}{\mathrm{\Delta t}}$ or $P=\frac{E}{T}$, where P is power, W is work, E is energy, and T is time. The unit of power is the watt (W), which is equivalent to joules per second (J/s).

How do you calculate the power consumed by a 100 W light bulb in one hour?

To calculate the power consumed by a 100 W light bulb in one hour, you use the formula $E=PT$. Here, P is 100 W and T is 1 hour, which is 3600 seconds. So, $E=100\times 3600=360000$ joules. Therefore, the light bulb consumes 360,000 joules of energy in one hour.

What is the unit of power and how is it related to energy and time?

The unit of power is the watt (W), which is equivalent to joules per second (J/s). Power is related to energy and time through the formula $P=\frac{E}{T}$, where P is power, E is energy, and T is time. This means that power measures how quickly energy is transferred or work is done over a period of time.

How do you calculate the average power delivered by a car engine during acceleration?

To calculate the average power delivered by a car engine during acceleration, you can use the formula $P=\frac{W}{\mathrm{\Delta t}}$. First, find the work done using the kinetic energy theorem: $W=\frac{1}{2}m{v}^2$. Then, divide by the time interval Δt. For example, if a 1300 kg car accelerates from 0 to 40 m/s in 7 seconds, the work done is $\frac{1}{2}1300\times {40}^2$, and the average power is $\frac{1}{2}1300\times {40}^2÷7$, which equals 149 kW.

What is the relationship between work, energy, and power?

Work, energy, and power are closely related concepts in physics. Work is the transfer of energy through force applied over a distance, expressed as $W=Fdcos\theta$. Energy is the capacity to do work. Power is the rate at which work is done or energy is transferred, given by $P=\frac{W}{\mathrm{\Delta t}}$ or $P=\frac{E}{T}$. Essentially, power quantifies how quickly work is performed or energy is used over time.