Question

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.

Answer 1

Answer:

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

Answer 2

The work done is 49.48 J.

The tool is acted upon by a force   F = $-axy^{2} j$

a = 3.00 $\frac{N}{m^{3} }$

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

Then the force F follows by the line:

F = $-ay^{3} j$,  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.

$W=\int\limits {F} \, ds$, here ds is the displacement element.

$W=\int\limits^{2.85}_0 {-ay^{3} } \, dy =-3*[y^{4}/4 ]^{2.85}_0$

W = -3 × $[(2.85)^{4}/4 ]$

W = -49.48 J

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