# Work done by a variable force

1. Mar 18, 2004

### cristina

A 3kg object moving along the x-axis has a velocity of 2.40m/s as it passes through the origin. It is acted on by a single force Fx that varies with x as shown...
a) find the work done by the force from x=0 to x = 2m
b) what is the kinetic energy of the object at x=2m?
c) what is the speed of the object at x=2m?
d) find the work done on the object from x=0 to x=4
e) what is the speed of the object at x =4m
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a) I know since the force is variable the work is equal to the area under the curve Fx-versus-x. I dont know wich value of F should I put in the Work integral equation S fx dx.

b) k = 1/2mv^2 =1/2(3kg)v^2
the force on the curve at x=2 is F = 1.9N can I put it the newton second law equation Ftotal = ma, 1.9N = 3kg x a ,
a = 1.9N/3kg = 0.63 m/s^2 and then since I got the acceleration I pluge it in v^2 = v + 2(a)(dleta x) = (2.40m/s)+2(0.63)(2m-0m)= 4.92
so v = 2.22m/s ?

d) before x=4m actually at x = 3m the curve go under the x-axis
I will integrate the part from x=0 to x=3 and then subtract from it the integration from x=3 to x = 4. but again I dont know what force should I use in the work integration fromula.

2. Mar 18, 2004

Well, without the graph of F vs. x, we're not going to be able to give you anything but general statements.

$$W = \int_{x_i}^{x_f} F(x) \,dx$$

which can be interpreted as the area under the curve of a graph of F vs. x. You're correct in that. If you have an equation for F(x), then you're going to put that the integral and evaluate the indefinite integral between the initial x point and the final x point, which are 0 and 2m respectively for part (a).

For (b), you're working way too hard. Work is the change in energy. You're to assume that all of the energy gained (or lost) by the applied force is transformed into (or taken away from) the kinetic energy. So all you have to do is calculate the kinetic energy at the origin and then add the work done between 0 and 2m (which you did in part (a)!).

(c) Work backwards using part (b).

(d) Same as part (a), except the inital point is 0 and the final point is 4m. If you want, you can split up the integral into two separates ones, one between 0 and 2m and the other between 2m and 4m, and use part (a).

(e) As part (c).

Hope this helps. If it's still giving you trouble, we'll be able to help more if you can give us a picture of the graph or at least describe it.

3. Mar 18, 2004

### cristina

Is the speed wrong in c)?

k = 1/2mv^2 =1/2(3kg)v^2
the force on the curve at x=2 is F = 1.9N can I put it the newton second law equation Ftotal = ma, 1.9N = 3kg x a ,
a = 1.9N/3kg = 0.63 m/s^2 and then since I got the acceleration I pluge it in v^2 = v + 2(a)(dleta x) = (2.40m/s)+2(0.63)(2m-0m)= 4.92
so v = 2.22m/s

I dont have an equation, If I had I would differentiat it as you said. all I have it a graph.

4. Mar 18, 2004

The problem with your approach is that the acceleration is not constant if the force is not constant, as in this case. The equation you used is valid only for constant acceleration (i.e. constant force).

As for finding the area under the curve, I can't really help you because I don't know what it looks like. If you're in high school, then I imagine you'll be able to figure it out by breaking it up into triangles and rectangles and using a bit of geometry, but who knows.

5. Mar 18, 2004

### cristina

I will break it into rectangles and triangles, this is a great idea actually. but I thought I can get the Force values and integrate it but force is not constant as you said. May I please ask you about the cirve at x = 3m the curve go under the x-axis so force is negative can I calculate the area of the from x=3 to x=4 and then subtact it from x=0 to x = 3 so I would have the calculated the work done on the object from x= 0 to x = 4, is it right the way I am thinking?

May you explain more on how to get the speed please?

6. Mar 18, 2004

When force is negative, then you're taking energy out of the system (not generally true, but it is in this case). Consequently, the change in enery (work) will be negative, so you'll just subtract the area below the curve from the area above the curve to get the total work.

As for the speed, calculate the kinetic energy at x = 0. The velocity at x = 0 is given. Now, the work is the change in energy. So if the energy started at the value you just calculated, then at x = 2m, it will be different by the amount of work done between x = 0 and x = 2m. Consequently, you can just add the work done to the original energy to get the energy at that point. Once you know the energy, you can calculate the velocity using E = 1/2mv^2.

$${v_f}^2 = {v_0}^2 + 2a\Delta x$$