There are instances where you won't need to evaluate the path integral to find work. Many cases allow us to simplify this path integral evaluation a bit. One such simplification arises when we assume the path to be linear and along (arbitrarily) the x-direction. This would lead to [math]W=\int{F_x}dx.[/math] This works because [math]\vec{A}\cdot\vec{B} = A_xB_x+A_yB_y+A_zB_z,[/math] which means that the integrand in the work expression could be written [math]\vec{F}\cdot\vec{ds}=F_xdx + F_ydy+F_zdz.[/math] Realize that [math]F_x =F\cos\theta[/math] if [math]\theta[/math] is the angle between the force and the x-axis.[br][br]A further simplification arises when the direction and magnitude of F is constant and the path is along a single axis. In this case you can remove [math]\vec{F}[/math] from the integral. Work then becomes [math]W=F_x\Delta x.[/math][br][br][color=#1e84cc][i]Example: Consider the work done by someone stretching a spring from equilibrium to a distance x away from equilibrium. This may be treated as a one-dimensional problem in which the force of the person is the opposite of the spring's force [math]F_x=kx[/math] which gives work [math]W=\int F_x\; dx = \int kx \; dx = k\tfrac{x^2}{2}[/math] since the integration limits are between zero and x. [/i][/color][br][br]One big message about the concept of work is that when an object is pushed (force) in the direction that it's moving, this amounts to positive work being done. According to the work-kinetic energy principle, such work increases the object's kinetic energy and speed. When something is pushed in a direction opposite to the direction in which it is moving, negative work is done on it and this leads to a decrease in the object's kinetic energy and speed. If pushed in a perpendicular direction to the motion, no work is done and the speed will remain constant. This can be seen from the dot product in the work definition where the cosine of a right angle gives zero.[br][br][color=#1e84cc][i]Example: A person pushed a shopping cart along an aisle for a distance of 30 m with a force of 10 newtons directed 30 degrees below horizontal. How much work did they do on the cart?[br][br]We need to find only the component of force along the direction of motion. This is [math]F_x = 10N \cos 30^o = 5\sqrt{3}N.[/math] Using this I get [math]W=5\sqrt{3}N(30m)\approx260J.[/math][/i][/color]

Let's consider another example that will require you to remember some things about friction and forces. [color=#1e84cc][i]Consider a child that sits on a sled on level ground in the snow. The combined mass of child and sled is m=20kg. The sled/snow interface has a coefficient of friction equal to [math]\mu_k=0.10.[/math] A parent pulls on a string attached to the front of the sled and drags the child along the level ground for a distance of [math]\Delta x=100m[/math] while [math]\Delta K=0.[/math] The string makes an angle [math]\theta = 30^{\circ}[/math] above the horizontal.[br][br]How much work is done by the parent and by friction during this process?[br][br]We must be careful here to realize that the normal force does not equal mg because part of the parent's force is directed upward and lowers the contact force. So we use Newton's 2nd law and find: [math]\sum\vec{F}=m\vec{a}.[/math] [br]This always reduces to [math]\sum F_x=ma_x,[/math] and [math]\sum F_y=ma_y.[/math] Here we find that in the y-direction, which I will assume is upward, we get [math]\sum{F_y}=F_{parent}\sin\theta+F_N-mg=0,[/math] and [math]\sum{F_x}=F_{parent}\cos\theta-\mu F_N=0[/math] in the x-direction.[br][br]Solving for the normal force in the first equation gives: [math]F_N=mg-F_{parent}\sin\theta,[/math] and plugging into the second equation we find: [math]F_{parent}\cos\theta - \mu(mg-F_{parent}\sin\theta) = 0.[/math] This leads to [math]F_{parent}=21.4N.[/math] So to get work done by the parent, we can use [math]W_{parent}=F_x\Delta x = 21.4N\cos 30^{\circ}100m = 1850 J.[/math] Since we know the kinetic energy isn't changing, the total work done on the sled by both the parent and by friction must equal zero. This means [/i][/color][math]W_{friction}=-1850J.[/math]

Recall that one reason we do all these work calculations is to be able to predict the change in the speed and associated kinetic energy of an object after work is done on it. If we consider the shopping cart example above. In general, we must be careful to consider all forces on an object, since they may all be doing work on the object. Once we know the net work done on an object by all forces acting on it, we can set that sum equal to the change in the object's kinetic energy, [math]\sum W=\Delta K.[/math] [br][br][i][color=#1e84cc]EXAMPLE: If the shopping cart in the previous example started from rest and had a mass of 40 kg and if no other horizontal forces were present, what should its speed have been after traveling the 30m distance?[/color][br][br][color=#1e84cc]We set [math]W=\Delta K.[/math] This leads to [math]260J = \frac{1}{2}m(v_f^2-v_i^2).[/math] Using [math]v_i = 0,[/math] this leads to [math]v_f = 3.6 \frac{m}{s}.[/math] We can be sure that the normal force and the gravitational forces don't do any work since they are directed orthogonally to the motion. The cosine term in the dot product makes the work zero then. [/color][/i][br][br][i]QUESTION: What if in reality the shopping cart did not pick up any speed, but rather moved along at some constant, non-zero speed along the aisle? What must have been going on?[br][br]If [math]\Delta K = 0[/math] (since v is constant) this means that [math]W_{net} = 0[/math] as well. This means that work is being done by something else besides the applied force of the shopper. We expect this other work to be done by friction. How much work must be done by friction? It has to be [math]-260J[/math] so that work by friction plus work by the shopper is zero.[/i]