3.2.1 Introduction to Work & Energy

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Presentation transcript:

3.2.1 Introduction to Work & Energy How does it do that? 3.2.1 Introduction to Work & Energy

Definitions energy: ABILITY TO DO WORK work: A CHANGE IN TOTAL ENERGY Unit: JOULE work: A CHANGE IN TOTAL ENERGY requires MOTION Unit: JOULE Equation

No, work is NOT being done ‘on the object’ unless it moves. Example #1 – No Motion A student is trying to push a heavy crate across the floor, but it does not move. Is work being done ‘on the crate’? No, work is NOT being done ‘on the object’ unless it moves.

Example #1 – No Motion But it feels to the student like energy is being used -- what’s going on? Energy IS being expended and work IS being done, just not ‘on the object’. Work is being done inside the student’s body.

Example #2 – Flat Ground A box is pushed with a force of 100 newtons across a frictionless surface for a distance of 10 meters. How much work is done on the box? 100 N 10 m

Example #2 – Flat Ground A box is pushed with a force of 100 newtons across a frictionless surface for a distance of 10 meters. How much work is done on the box? 100 N 10 m

Example #2 – Flat Ground A box is pushed with a force of 100 newtons across a frictionless surface for a distance of 10 meters. How much work is done on the box? W = Fd = ΔET W = (100N)(10m) W = 1000 J 100 N 10 m

Example #3 – Incline A box is pulled up a 3.0 meter long plane that is inclined at 20° using a force of 10 newtons. How much work is done in moving the box? 10 N 3.0 m 20°

Force is same direction as motion SO W = Fd applies Example #3 – Incline A box is pulled up a 3.0 meter long plane that is inclined at 20° using a force of 10 newtons. How much work is done in moving the box? Force is same direction as motion SO W = Fd applies 10 N 3.0 m 20°

Example #3 – Incline A box is pulled up a 3.0 meter long plane that is inclined at 20° using a force of 10 newtons. How much work is done in moving the box? W = Fd = ΔET W = (10N)(3.0m) W = 30 J 10 N 3.0 m 20°

Example #4 – Force on an Angle A sled is pulled a horizontal distance of 4.0 meters across a frictionless surface using a force of 15 newtons at an angle of 30° above the horizontal. How much work is done on the sled? 15 N 30° 4.0 m

Example #4 – Force on an Angle A sled is pulled a horizontal distance of 4.0 meters across a frictionless surface using a force of 15 newtons at an angle of 30° above the horizontal. How much work is done on the sled? Force is DIFFERENT direction than motion… We need to find the horizontal component of the force 15 N Horizontal Component 30° 4.0 m

Example #4 – Force on an Angle A sled is pulled a horizontal distance of 4.0 meters across a frictionless surface using a force of 15 newtons at an angle of 30° above the horizontal. How much work is done on the sled? (1) Ax = Fx = Fcos θ = (15N) cos(30°) = 12.99 N (2) W = Fxd =ΔET W = (12.99 N)(4.0m) W = 52 J 15 N 30° 4.0 m

End of 3.2.1 - PRACTICE

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