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PHYS2001_Ch. 6

# PHYS2001_Ch. 6 - Chapter 6 Work and Energy 6.1 Work done by...

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Chapter 6 – Work and Energy 6.1 Work done by a constant force Energy. What is it??? You all probably have an idea of what energy is. It comes in many forms: mechanical, chemical, heat, light, nuclear, etc. And energy comes in two types: Kinetic Energy – Energy of motion Potential Energy – Energy due to position Energy is governed by certain laws which we will learn in this chapter. And we will find that solving certain problems with an energy approach is often much easier than using Newton’s Laws. But both of these very different approaches are linked through a common concept, and that is the concept of Work . We all have our own ideas about what work means to us, but in physics, work is very explicitly defined. It is defined in terms of Force and Displacement . Work done on an object by a constant force : d F d = W (*Note: Do not confuse work with weight!)

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d F d = W In words, Work is equal to Force times distance. In the equation for work, F d is not necessarily the total force on the object, but it is the net force (or component of the force) along the direction of motion, i.e. either in the same direction as the displacement, d , or in the opposite direction of the displacement, d . Work is a scalar , not a vector, but it can be either positive or negative. Work is positive if the force and displacement are in the same direction. Work is negative if the force and displacement are in opposite directions. Units? [ ] ent) (displacem Force) ( × [ ] m N [ ] J (Joules) Now that we know what work is, we can more explicitly define Energy: Energy is the ability to do work.
Qualitatively, how much work am I doing on the medicine ball? 1. Quite a lot. 2. None. Clicker Question d F d = W 0 0 =

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Examples : A large crate slides on a horizontal frictionless floor from left to right for a distance x . A constant force F is applied to the crate in the different cases as shown. Calculate the work done on the crate in each case. x x x x 30 o F F F F 1 4 3 2 d F d = W . 1 Fx = d F d = W . 2 x F x = d F d = W . 3 Fx - = d F d = W . 4 0 = F ) )( cos ( x F θ = Negative since the force is in the opposite direction of the displacement. Zero! No work done since there is no component of F along the displacement.
Example : A 50-kg crate is pulled across a frictionless horizontal warehouse floor by a constant force of 750 N. The force is directed at 25 o above the horizontal. How much work is done on the crate by the pulling force, if the crate is moved a distance of 15.2 m? 25 o 15.2 m F = 750 N x y The crate slides in the x -direction, so that is the direction of the displacement. Therefore, we need the total force in the x -direction to calculate the work. F d F d = W d F x = d F o = 25 cos N 330 , 10 ) 2 . 15 )( 25 )(cos 750 ( = = o

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Two forces act on the box shown in the drawing, causing it to move across the floor. The two force vectors are drawn to scale. Which force does more work on the box?
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PHYS2001_Ch. 6 - Chapter 6 Work and Energy 6.1 Work done by...

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