Calculate the work done on the tool by F⃗ if the tool is first moved out along the x-axis to the point x= 2.95 m , y=0 and then moved parallel to the y-axis to x= 2.95 m , y= 2.95 m .

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.

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

the amount of current in the coil

answer would be letter choice (a),   a ball resting on a shelf. would be another example of potential energy   >   stored energy > not moving.

a ball spinning on a finger   >   kinetic energy > moving energy

a ball rolling on the ground   >   kinetic energy   > moving energy

a   ball flying through the air   > kinetic energy   >   moving energy

hope that                                         : )