Question

Determine the work done by the force Fx=(6x-4)N as the object moves from x1=0 to x2=3 m.

Answer 1

15J

Step-by-step explanation:

Let's assume force and motion have the same direction, so the dot product between the two reduces to the product of magnitudes.

If you studied calculus:
`w= \int \limits^3_0 6x-4dx`, by definition. Some "easy" math, that's
`(3x^2-4x)\limits^3_0=27-12 - 0 = 15 J`

Else. we have to take the long way. if the force was constant, and you plotted magnitude of the force vs position in a graph, you would get an horizontal line, and your work would be the area between the line and the x axis. And there is no reason why it shouldn't be the same for a variable force - if you think about it, it's the same way they justify the position formula for an accelerated motion. Let's graph it and start debating - sorry for my paint skills. We can easily say that in the green triangle the force is opposing the motion, so its work is negative, while in the blue area force and displacement go in the same direction, thus work is positive. Now it's just to calculate the area of two triangles

For the green one, it's a right triangle with side lengths 4 and 2/3, whose area is
`\frac12(4)(\frac23) = \frac43`.

For the blue triangle, it's again a right triangle with side lengths 14 and 3-2/3= 7/3, whose area is
`\frac12(14)(\frac73) = \frac{49}3`

Our total work is the difference between the two, or
`w= \frac{49}3-\frac43 = \frac{45}3 = 15`