Question

A cutting tool several forces acting on it. One force is F=-axy^2 j , a force in the negative y-direction whose magnitude depends on the position of the tool. The constant is a= 2.85 . Consider the displacement of the tool from the origin to the point x=2.40 , y=2.40 .

Calculate the work done on the tool by F if this displacement is along the straight line y=x that connects these two points.

Calculate the work done on the tool by F if the tool is first moved out along the x-axis to the point x=2.40, y=0 , and then moved parallel to the y-axis to x=2.40

Answer 1

I know the integral of non constant force is work but not sure how to play with the given force to get work

Answer 2

do you know how to do a line integral?

Answer 3

parameterize the path the object moves along as\[\vec r(t)=g(t)\hat i+h(t)\hat j~~~~\text{ for }~~~~~a\le t\le b\]\[W=\int_a^b f(g(t),h(t))\|\vec r'(t)\|dt\]

Answer 4

Im not sure what to do

Answer 5

do you know how to parameterize the line segment from y=x from x=0 to x=2.4 ?

Answer 6

never heard of that word

Answer 7

okay I see what they want...

Answer 8

I didn't realize it had two separate motions... first along x, then along y
y=x
hence F=-axy^2 j=-ax^3 j=-ay^3 j
now remember that work is done by force only along the direction of motion
therefor when the object moves along x (the i direction) no work is done, and when the object moves along the y direction the work done is the integral of the force with the path it travels

Answer 9

How do you know that: object moves along x (the i direction) no work is done, and when the object moves along the y direction ((in this case))

Answer 10

I don't know what level physics you are in, so I don't know what explanation will make sense to you
do you know what a dot product is?

Answer 11

Yes

Answer 12

(physics with calc 1)

Answer 13

Well, a better formula for work is\[W=\vec F\cdot\vec d=Fd\cos\theta\]The force only acts in the j-direction (along x). The angle between the force and the x-axis is pi/2, so \(W=Fd\cos\theta=0\)
along the x-axis the angle between is 0 degree, the cosine of 0 is 1, is for that portion of the motion \(W=Fd\)

Answer 14

singe the force is proportional to the y-coordinate though, you must integrate along the y-axis\[W=\int_a^bFdy\]

Answer 15

ohh because the j direction is given ok got it..cant believe i missed that :s

Answer 16

glad it makes sense :)