A cutting tool under microprocessor control has several forces acting on it. One force is
→F=−αx(y^2) (^j) , a force in the negative y-direction whose magnitude depends on the position of the tool. The constant is
α= 2.70 N/m^3. Consider the displacement of the tool from the origin to the point x= 2.95m, y=
2.95m. Calculate the work done on the tool by →F if the tool is first moved out along the x-axis to the point x = 2.95m, y= 0m and then moved parallel to the y-axis to
x= 2. 95m , y= 2.95m.
Answer: W = 59.56 Joules
Explanation:
We know that, dW = f(ds)
Therefore, W= ∫→F(→ds)
Since force is only along y-axis ,work done along x- axis is zero and thus work done in moving from
(2.95,0) to (2.95,2.95) is:
W = (2.95,0)∫ - αx(y^2) dy(cos 180°)
Cos 180° = - 1 and x =2.95
So, W= (2.95,0)∫ 2.95α(y^2) dy
α = 2.7, so W = (2.95,0)∫ (2.95 x 2.7)(y^2) dy
W = (2.95,0) ∫ (6.96)(y^2) dy
Integrating this with the boundaries (2.95,0), we get;
W = 6.96 ( (2.95^(3))/3) - 0)
W = 6.96 x 8.557 = 59.56 Joules
