Ch. 07 - Work and Energy

Giancoli Douglas - Physics for Scientists and Engineers 5th edition

Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Not the one you use?Change textbook

Giancoli Douglas 5th edition

Ch. 07 - Work and Energy

Problem 54

Chapter 7, Problem 54

At room temperature, an oxygen molecule, with mass of 5.31 x 10⁻²⁶ kg, typically has a kinetic energy of about 6.21 x 10⁻²¹ J . How fast is it moving?

Verified step by step guidance

Step 1: Recall the formula for the kinetic energy of an object:

K = \frac{1}{2} m v^{2}

, where

K

is the kinetic energy,

m

is the mass, and

v

is the velocity.

Step 2: Rearrange the formula to solve for velocity

v

. Start by multiplying both sides of the equation by 2 to eliminate the fraction:

2 K = m v^{2}

Step 3: Divide both sides of the equation by

m

to isolate

v^{2}

v^{2} = \frac{2}{K} / m

Step 4: Take the square root of both sides to solve for

v

v = \sqrt{\frac{2}{K} / m}

Step 5: Substitute the given values for

K

(6.21 × 10⁻²¹ J) and

m

(5.31 × 10⁻²⁶ kg) into the formula and compute the result to find the velocity.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Kinetic Energy

Kinetic energy is the energy an object possesses due to its motion, calculated using the formula KE = 1/2 mv², where m is mass and v is velocity. In this context, the kinetic energy of the oxygen molecule is given, which allows us to determine its speed by rearranging the formula to solve for v.

Mass

Mass is a measure of the amount of matter in an object, typically expressed in kilograms. In this question, the mass of the oxygen molecule is provided as 5.31 x 10⁻²⁶ kg, which is essential for calculating its velocity using the kinetic energy formula.

Velocity

Velocity is a vector quantity that describes the speed of an object in a specific direction. In this scenario, we need to calculate the velocity of the oxygen molecule based on its kinetic energy and mass, which will provide insight into its motion at room temperature.

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