Answer:
180 fb*lb
45 ft*lb
Step-by-step explanation:
We have that the work is equal to:
W = F * d
but when the force is constant and in this case, it is changing.
therefore it would be:
![W = \int\limits^b_ a {F(x)} \, dx](https://tex.z-dn.net/?f=W%20%3D%20%5Cint%5Climits%5Eb_%20a%20%7BF%28x%29%7D%20%5C%2C%20dx)
Where a = 0 and b = 30.
F (x) = 0.4 * x
Therefore, we replace and we would be left with:
![W = \int\limits^b_a {0.4*x} \, dx](https://tex.z-dn.net/?f=W%20%3D%20%5Cint%5Climits%5Eb_a%20%7B0.4%2Ax%7D%20%5C%2C%20dx)
We integrate and we have:
W = 0.4 / 2 * x ^ 2
W = 0.2 * (x ^ 2) from 0 to 30, we replace:
W = 0.2 * (30 ^ 2) - 0.2 * (0 ^ 2)
W = 180 ft * lb
Now in the second part it is the same, but the integral would be from 0 to 15.
we replace:
W = 0.2 * (15 ^ 2) - 0.2 * (0 ^ 2)
W = 45 ft * lb