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Standard XII

Physics

Work Done, Varying Force and Path

A force F⃗ ...

Question

A force $\to F = - K (y^i + x^j)$ , where $K$ is a positive constant, acts on a particle moving in the X-Y plane. Starting from the origin, the particle is taken along the positive X-axis to a point $(a, 0)$ and then parallel to the y-axis to the point $(a, a)$ . Calculate the total work done by the force on the particle.

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Solution

F = -k(yi+xj)
W = F.dr              (dot product of force & displacement)
displacement vector(dr) is of line joining initial & final position vectors...
W = -k(yi+xj).(dxi+dyj)
dW = -k(ydx + xdy)
dW = -kd(xy)  (exact)
integrating both sides ,
W = -k [xy] lim from (0,0) to (a,a)
W  = -ka²
Answer

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F = -k(yi+xj)W = F.dr (dot product of force & displacement)displacement vector(dr) is of line joining initial & final position vectors...W = -k(yi+xj).(dxi+dyj)dW = -k(ydx + xdy)dW = -kd(xy) (exact)integrating both sides ,W = -k [xy] lim from (0,0) to (a,a)W = -ka2Answer

F = -k(yi+xj)
W = F.dr (dot product of force & displacement)
displacement vector(dr) is of line joining initial & final position vectors...
W = -k(yi+xj).(dxi+dyj)
dW = -k(ydx + xdy)
dW = -kd(xy) (exact)
integrating both sides ,
W = -k [xy] lim from (0,0) to (a,a)
W = -ka²
Answer