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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 Δx, the work done by the force on the object is


where φ is the angle between the direction of the force F and the direction of the displacement âx. The units for work are [kg m2/s2 = Nm = Joule = J].


The total work done on an object is


where Fnet is the magnitude of the net force and φ is the angle between Fnet and âx.

Keep in mind that if the force is in the same direction as the motion, then cosφ = 1. If the force exactly opposes the motion, then cosφ = –1. If the force acts perpendicular to the motion, then cosφ = 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.


A woman is pushing a cart of mass m slowly at constant speed up an incline which makes an angle θ with the horizontal. The cart goes from the floor level to a height h.
  1. How much work does the woman do on the cart in terms of mgh, and θ?
  2. What is the total work done on the cart?
  1. First, we DRAW A DIAGRAM (Figure 9-1).

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Figure 9-1

We need to find FW and Δx. We choose a "horizontal" and "vertical" and resolve the gravity vector into components (Figure 9-2).


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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
(Fnet)x = 0
From Figure 9-2 we obtain
(Fnet)x FW – mg sinθ
Combining these equations gives us
Fw – mgsinθ = 0
Fw = mgsinθ
We can find Δx by trigonometry. If we look at the large triangle in Figure 9-2, then we have
sinθ = hx
Δxsinθ = h
Δx = h/sinθ
The vectors FW and Δx point in the same direction, so cosφ = 1. Thus the work done by the woman on the cart is


Ww = mgh
This is the answer to question a.
But what is this? The quantity θ 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.

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Figure 9-3

  1. What is the total work done? Well, since the speed and direction are constant, the acceleration is zero, and Fnet = 0, so that
Wtot = 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 –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).


How much work is done by the gravity of the Sun on the Earth in one day?
G = 6.67 x 10–11 m3/kg s2,
MSun = 2 x 1030 kg,
MEarth = 6 x 1024 kg,
distance from Earth to Sun = 1.5 x 1011 m.

First, we DRAW A DIAGRAM (Figure 9-4). Once we draw the diagram, the answer is clear. The vector Fgrav is perpendicular to the displacement Δx, so we have cosφ = 0 and Wtot = 0.


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Figure 9-4


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