An object moves along a straight track from the point (1,-2,1) to the point

Question

An object moves along a straight track from the point (1,−2,1) to the point (0,−7,5). The only force acting on it is a constant F=2i−3j−5k newtons. Find the work done if the distance is measured in meters.

Answer 1

Recall that work along a trajectory can be written as a line integral

∫F•dr = w.

We need a few things to be able to carry out the integration:

1. vector expression for F
2. parametric equation for the trajectory connecting the two given endpoints (this will be r(t))
3. derivative of r so we can carry out the dot product and integrate

The first item on our list is simple; we were already given F=[2, -3, -5] in newtons. We can move on for now.

 

We need to parameterize the straight line connecting the two given endpoints. The parameter I'll use here is t. Notice that if we subtract the coordinates of the initial point from the terminal point, we can construct a difference vector that lives along this line:

[(0-1), (-7 - -2), (5-1)] ⇒ [-1, -5, 4] 

Now comes the step that relies on our intuition (and creativity). When we parameterize a trajectory curve, we introduce the parameter variable (t), so we can say, completely arbitrarily, that the parameter is restricted to some interval. In other words, a≤t≤b, where a and b are the two endpoints of the interval. When t=a, we should expect the curve to be at its starting point. Likewise, when t=b we expect the curve to be at its ending point. A really easy interval for straight lines is 0≤t≤1.

 

We can parameterize the curve connecting the two given points as

r(t)= [1, -2, 1]+[-1, -5, 4]t = [(1-t), (-2-5t), (1+4t)].

Make sure you understand where this comes from. We made a vector out of the initial position [1, -2, 1], and then added on the difference vector multiplied by t. When t=0, r(t) is just the initial position, like we wanted. When t=1, we add the full difference vector to the initial position vector, and obtain the coordinates of the second endpoint. While t is in between 0 and 1, we are somewhere on the line connecting the two points.

 

Moving on to the third step, we want to modify our work integral into something a little more useful:

w = ∫F•dr = ∫F•(dr/dt) dt

This could seem a bit trivial, but there are important consequences. The F*dr form of the integral says we're computing work as {force}*{displacement}. We know from physics that displacement is the same as {velocity}*{time}, so we make that substitution. For a given trajectory r(t), the associated velocity is just dr/dt. The integral is now taken over dt instead of dr, and the limits of integration are from t=0 to t=1 instead of r=(1,-2,1) to r=(0,-7,5).

 

So what is dr/dt? We take the derivative of each individual component of r:

d/dt ([(1-t), (-2-5t), (1+4t)]) = [d/dt(1-t), d/dt(-2-5t), d/dt(1+4t)] = [-1, -5, 4].

 

Plugging everything we have into the integral, we see it looks like this:

w = ∫[2, -3, -5]•[-1, -5, 4] dt;      t∈[0,1]

The dot product evaluates to -7 and 

w = ∫(-7)dt = -7.