A cutting tool under microprocessor control has several forces acting on it. One force is F⃗ =−αxy2j^, a force in the negative y-direction whose magnitude depends on the position of the tool. The constant is α = 3.00 N/m3 . Consider the displacement of the tool from the origin to the point x= 2.85 m , y= 2.85 m .a. Calculate the work done on the tool by F⃗ if this displacement is along the straight line y=x that connects these two points.

The work done is 49.48 J.

The tool is acted upon by a force   F =

a = 3.00

Now the tool is displaced along the straight line y = x

Then the force F follows by the line:

F = ,  putting x = y so that it satisfies the given straight line equation y = x

So we need only y-coordinate to calculate the work done W.

, here ds is the displacement element.

W = -3 ×

W = -49.48 J

Learn more about work done:

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Explanation:

Given that

F=-αxy²j

α=3N/m³

Displacement from x=2.85 to y=2.85

Since, y=x

Then dy=dx

ds=dx i + dy j

a. Work done

W=∫ F.ds from (0,0) to (2.85,2.85)

W=∫(-αxy²j).(dx i + dy j) (0,0)—(2.85,2.85)

Note, i.i=j.j=1 i.j=j.i=0

W=∫-αxy²dy from (0,0)—(2.85,2.85)

W=-α∫xy²dy from (0,0)—(2.85,2.85)

x=y

W=-α∫y³dy from (0,0)—(2.85,2.85)

W=-αy⁴/4. from (0,0)—(2.85,2.85)

W=-3[(2.85)⁴/4-0]

W=-49.48J

The work done on the tool by the force is -49.48J

statement #3 on andre"s list is false.

it's when something is going so fast it makes a loud noise, breaking the sound barrier, for example: a jet makes a loud noise, not only cause of the jet but because they reached a point where they went so fast that you can year the noise from various parts, kinda like a police car, lets say the car was going   so fast that you hear the siren of the cop car right next to you, but the car is probably a mile a way, it's cos the siren was next to you a few seconds if you understand me