# Work

Work is a measure of the energy flowing into an object or system due to a force on it. When a force*F*(which can be due to a pair of hands, a rope, gravity, or anything) acts on an object which moves a distance Î”

*x*, the

*work*done by the force on the object is

â€¦(1) |

where

*Ï†*is the angle between the direction of the force*F*and the direction of the displacement â*x*. The units for work are [kg m^{2}/s^{2}= Nm = Joule = J].

The

*total work*done on an object isâ€¦.(2) |

where

Keep in mind that if the force is in the same direction as the motion, then cos

*F*_{net }is the magnitude of the net force and*Ï†*is the angle between*F*_{net }and â*x*.Keep in mind that if the force is in the same direction as the motion, then cos

*Ï†*= 1. If the force exactly opposes the motion, then cos*Ï†*= â€“1. If the force acts perpendicular to the motion, then cos*Ï†*= 0. You should know these angles without pausing to think about them.

**Hint:**Whenever a problem on the MCAT mentions a force and a distance, you should think "WORK!" and write down the equation for work. Even if work and energy are not mentioned, it is probably a key idea for understanding at least one of the problems.

Example-1

A woman is pushing a cart of mass

*m*slowly at constant speed up an incline which makes an angle*Î¸*with the horizontal. The cart goes from the floor level to a height*h.*- How much work does the woman do on the cart in terms of
*m*,*g*,*h*, and*Î¸*? - What is the total work done on the cart?

Solution

- First, we DRAW A DIAGRAM (Figure 9-1).

**Figure 9-1**

We need to find *F*_{W }and Î”*x*. We choose a "horizontal" and "vertical" and resolve the gravity vector into components (Figure 9-2).

**Figure 9-2**

First we look at the "horizontal" components. The words "constant speed" and "straight path" imply that the cart's acceleration is zero, and the net force on it is zero, so we write

(

From Figure 9-2 we obtain
*F*_{net})_{x }= 0*(F*

_{net})

_{x }=

*F*â€“

_{W}*mg*sin

*Î¸*

*F*

_{w}â€“

*mg*sin

*Î¸*= 0

*F*

_{w}=

*mg*sin

*Î¸*

*x*by trigonometry. If we look at the large triangle in Figure 9-2, then we have

sin
Î”
Î”

The vectors *Î¸*=*h*/Î”*x**x*sin*Î¸*=*h**x*=*h*/sin*Î¸**F*

_{W }and Î”

*x*point in the same direction, so cos

*Ï†*= 1. Thus the work done by the woman on the cart is

*W*

_{w}=

*mgh*

This is the answer to question a.

But what is this? The quantity

*Î¸*dropped out of the equation (!). It takes the same energy to go a long way up a shallow incline as it does to go a short way up a steep incline. In fact:

The work done depends only on the height climb h from the begnning to end and not at all on the path between the two. |

The energy a woman requires to push a cart from one point to a point which is height

*h*higher is

*mgh*, even if the path is complicated. See Figure 9-3.

**Figure 9-3**

- What is the total work done? Well, since the speed and direction are constant, the acceleration is zero, and
*F*_{net }= 0, so that

*W*

_{tot}= 0

What? The poor woman works from dawn till dusk, and the total work done is zero? Where did the energy go? Well, if we were to figure out the work done by gravity, we would find that it comes to â€“

*mgh*, so the work done by gravity cancels the work done by the woman. It seems like a sad story, perhaps, but this will not be the final word on gravity (see Section D).

Example-2

How much work is done by the gravity of the Sun on the Earth in one day?

*G*= 6.67 x 10

^{â€“11}m

^{3}/kg s

^{2},

*M*

_{Sun }= 2 x 10

^{30}kg,

*M*

_{Earth }= 6 x 10

^{24}kg,

distance from Earth to Sun = 1.5 x 10

^{11}m.Solution

First, we DRAW A DIAGRAM (Figure 9-4). Once we draw the diagram, the answer is clear. The vector *F*_{grav }is perpendicular to the displacement Î”*x*, so we have cos*Ï†* = 0 and *W*_{tot }= 0.

**Figure 9-4**