Physics SL
Physics SL
5
Chapters
329
Notes
Theme A - Space, Time & Motion
Theme A - Space, Time & Motion
Theme B - The Particulate Nature Of Matter
Theme B - The Particulate Nature Of Matter
Theme C - Wave Behaviour
Theme C - Wave Behaviour
Theme D - Fields
Theme D - Fields
Theme E - Nuclear & Quantum Physics
Theme E - Nuclear & Quantum Physics
IB Resources
Theme A - Space, Time & Motion
Physics SL
Physics SL

Theme A - Space, Time & Motion

Coriolis' Insight: Understanding Work Done & Energy Transfer

Word Count Emoji
677 words
Reading Time Emoji
4 mins read
Updated at Emoji
Last edited on 5th Nov 2024

Table of content

Work done in physics

Work, in the context of physics, is the transfer of energy that occurs when a force moves an object over a distance. It was Gaspard-Gustave Coriolis, in 1826, who described this concept as he was studying engineering aspects of water extraction from a flooded mine.

 

He observed the energy transfer happening when steam engines pumped the water from the mine's bottom to the surface. This transfer of energy happened due to the engines exerting a force on a certain mass of water and lifting it vertically. Therefore, we can write this as

 

Work done (J) = Force exerted (N) x Distance moved in the direction of the force (m)

 

For example, when a 5N water weight is lifted vertically through 150m, the work done by the engine is 5N * 150m = 750J.

 

Work When Force and Distance are Not in the Same Direction: In some situations, the force and the distance moved are not in the same direction. For instance, consider a sand yacht where the wind force and the yacht's displacement are at an angle θ to each other. In such cases, we can use the component of force in the direction of movement, calculated as F cos θ. Hence, work done = F cos θ x s.

 

Work Done Against a Resistive Force: Work also happens when there's a resistive force, like friction. For instance, when you're pushing a box at a constant speed on a straight line, you must overcome friction. The force that beats this friction may not act in the direction of movement. So, the work done is: force acting x distance traveled x cos θ.

Practical examples

Let's explore some worked examples to understand these concepts bette.

 Microlight Aircraft

  • A microlight aircraft engine has a thrust (driving force) of 3.5 x 10³ N. When the aircraft travels a distance of 15km, the work done by the thrust is calculated as 3500N (force) x 15,000m (distance) = 5.3 x 10⁷ J = 53MJ.

Moving a Large Box

  • If you pull a large box 8.5m along a rough horizontal surface by a force of 55N that acts at 50° to the horizontal, you can calculate the work done by first finding the component of force in the direction of travel (55N x cos 50° = 35.4N) and then multiplying this by the distance traveled - 35.4N x 8.5m = 301J.

Cart Rolling Down a Ramp

  • A cart rolls down a ramp at an angle of 25° with the horizontal, with a horizontal force of 6.0N. The cart moves 0.75m down the ramp. To find the work done by the horizontal force, we first calculate the angle between the force and the direction of motion (180° - 25° = 155°). Then, we multiply the force, distance, and cos of this angle: 6.0N x 0.75m x cos 155° = -4.1J. The negative sign indicates that the horizontal force has a component acting against the cart's displacement - the force opposes the motion.

Car Moving at Constant Speed

  • If a car moves at a constant speed of 50km/h on a horizontal road and the work done by the driving force of the car in one minute is 190kJ, you can calculate - a. The distance traveled by the car in one minute, b. The magnitude of the resistive force acting on the car, and c. The work done by the resistive force.

By understanding these principles and solving similar problems, you can get a solid grasp of the concept of work in physics, and how it applies to different situations in the real world.

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IB Resources
Theme A - Space, Time & Motion
Physics SL
Physics SL

Theme A - Space, Time & Motion

Coriolis' Insight: Understanding Work Done & Energy Transfer

Word Count Emoji
677 words
Reading Time Emoji
4 mins read
Updated at Emoji
Last edited on 5th Nov 2024

Table of content

Work done in physics

Work, in the context of physics, is the transfer of energy that occurs when a force moves an object over a distance. It was Gaspard-Gustave Coriolis, in 1826, who described this concept as he was studying engineering aspects of water extraction from a flooded mine.

 

He observed the energy transfer happening when steam engines pumped the water from the mine's bottom to the surface. This transfer of energy happened due to the engines exerting a force on a certain mass of water and lifting it vertically. Therefore, we can write this as

 

Work done (J) = Force exerted (N) x Distance moved in the direction of the force (m)

 

For example, when a 5N water weight is lifted vertically through 150m, the work done by the engine is 5N * 150m = 750J.

 

Work When Force and Distance are Not in the Same Direction: In some situations, the force and the distance moved are not in the same direction. For instance, consider a sand yacht where the wind force and the yacht's displacement are at an angle θ to each other. In such cases, we can use the component of force in the direction of movement, calculated as F cos θ. Hence, work done = F cos θ x s.

 

Work Done Against a Resistive Force: Work also happens when there's a resistive force, like friction. For instance, when you're pushing a box at a constant speed on a straight line, you must overcome friction. The force that beats this friction may not act in the direction of movement. So, the work done is: force acting x distance traveled x cos θ.

Practical examples

Let's explore some worked examples to understand these concepts bette.

 Microlight Aircraft

  • A microlight aircraft engine has a thrust (driving force) of 3.5 x 10³ N. When the aircraft travels a distance of 15km, the work done by the thrust is calculated as 3500N (force) x 15,000m (distance) = 5.3 x 10⁷ J = 53MJ.

Moving a Large Box

  • If you pull a large box 8.5m along a rough horizontal surface by a force of 55N that acts at 50° to the horizontal, you can calculate the work done by first finding the component of force in the direction of travel (55N x cos 50° = 35.4N) and then multiplying this by the distance traveled - 35.4N x 8.5m = 301J.

Cart Rolling Down a Ramp

  • A cart rolls down a ramp at an angle of 25° with the horizontal, with a horizontal force of 6.0N. The cart moves 0.75m down the ramp. To find the work done by the horizontal force, we first calculate the angle between the force and the direction of motion (180° - 25° = 155°). Then, we multiply the force, distance, and cos of this angle: 6.0N x 0.75m x cos 155° = -4.1J. The negative sign indicates that the horizontal force has a component acting against the cart's displacement - the force opposes the motion.

Car Moving at Constant Speed

  • If a car moves at a constant speed of 50km/h on a horizontal road and the work done by the driving force of the car in one minute is 190kJ, you can calculate - a. The distance traveled by the car in one minute, b. The magnitude of the resistive force acting on the car, and c. The work done by the resistive force.

By understanding these principles and solving similar problems, you can get a solid grasp of the concept of work in physics, and how it applies to different situations in the real world.

Unlock the Full Content! File Is Locked Emoji

Dive deeper and gain exclusive access to premium files of Physics SL. Subscribe now and get closer to that 45 🌟