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From MyMCAT

Contents 1 Introduction 2 Work 3 The importance of the Path (Distance) 4 The importance of Cos(θ)

Introduction

In the previous section, Basic Kinematics, we saw how translational motion can be described in terms of displacement, velocity, and acceleration. In Newton's Laws Introduction we saw how translational motion can also be described using the previous terms along with mass and force. Now, we will again try to describe the motion of objects, but this time look at it in terms of kinetic energy, potential energy, and work. Unlike the previous sections, which all dealt with vectors, work and energy are scalars and thus it provides an alternative and often easier means to solve motion questions.

Work

Its common in every day conversation to say that doing a certain physical task is a lot of "work", but what does this term actually refer to mathematically? Consider two situations involving pushing a block. In the first, lets say the block is only 5 kg and we push it 10 meters. In the second situation, the block is 50 kg and the distance we push it 30 meters.

From our intuition of what work is, the second scenario is surely more "work", but why? What has changed? The block is heavier AND the distance we have to push it is longer. The formal definition of work takes both of these concepts into account,

In standard notation, the units N*m (Newton-meters) can describe work, however it has been given a more meaningful unit, that of energy in Joules (1J = 1Nm). Notice how the energy or work required to push a block will increase if we have to either apply more force or push it further. Pay attention not to forget the cos(θ) term as it determines the relevant amount of force being applied. If one wished to push a block horizontally then they should be applying the force in the horizontal direction as pushing downwards on it from above would be futile. cos(θ) accounts for this by factoring how much force is actually being applied in the direction of the movement. As expected, applying a force perpendicular to the direction of movement will result in NO work being done (as cos(90) equals 0).

The importance of the Path (Distance)

Pay special attention to the term d in the formula W=Fd. d is the distance of the total path traveled, not displacement. As the object is moved, work is always being done. Thus, regardless of where the path ends, were it be back at the initial position, work has still been done along the path, and so it is the length of the path that matters to calculating work, not the displacement from the initial position.

The importance of Cos(θ)

A child moves his toy car, which weighs 500 grams, a total of 20 cm from left to right in the air horizontally. How much work does gravity do during this task?

Before we jump to the answer, lets think about this. W=Fdcosθ. We know what the distance is, its 20cm, but what about the force? Is gravity the force that is pushing the toy car from left to right? No, it is the child's hand. What is gravity doing? Gravity isn't even in the same direction as the movement, it is in fact 90 degrees off from the direction of movement, which means that the work done by gravity (if we assume that it is the force which were causing the movement) is gravity*distance*cos(90), which is zero. Why? Because gravity isn't acting in the direction of the movement so it can't be what is causing the movement!

It can be confusing to consider holding a heavy box stationary in the air as requiring no work. But this is because we are referring to work as a mechanical physics term. In reality obviously, our bodies expend a lot of energy keeping our muscles tight to hold the box in the air.

When we apply a force to an object directly in the direction of the movement, the angle between them is zero. As such, Cos(0) = 1 and thus the maximum possible amount of force is applied in that direction. As we increase the angle however, we are reducing the effective force being applied to the direction of movement, and as such we expect the amount of work done to also be reduced. Finally, when we reach 90 degrees, the force we are applying has no component in the direction of the movement we are trying to achieve and as such, the work done, no matter how hard we push, is zero (because cos(90) is always 0).

1. Which of the following is NOT a correct unit for work?

	J
	N*m
	kg*m2/s2
	Kg*m/s
→	Work, by definition is a calculation of mechanical energy, and although we have just glanced at it, energy can be measured in Joules (J). Work is also equal to F*d, thus it can be measured in Newton-meters. Furthermore, Newtons, the measurement of force, can be replaced with kg*m/s2 (from F=ma). Therefore only D does not work as a meaningful measure of work.

2. How much work is performed in pushing a 2kg block 5 meters to the left and then 10 meters to the right, if friction always opposes the movement by 10N.

	150J
	100J
	50J
	There is not enough information to answer the question.
→	Work = force * distance. To overcome friction, we will need to push with 10N. Furthermore, the distance we actually push is 15m. In the case of work, we are concerned with the complete distance the block travels NOT the displacement because work is being done the WHOLE time in resisting friction.

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This is way more hpelful than anything else I've looked at.