#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.

Click to Chat

1800-1023-196

+91 7353221155

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

# A force F=-k(yi+xj) where k is a positive constant acts on a particle moving in the xy plane. Starting from the origin the particle is taken along the positive x axis to the point (a,0) and then parallel to y axis to the point (a,a). Find the total work done by the force F in the particle.

## 5 Answers

10 years ago

F = -k(xi+yj)

W = F.dr              (dot product of force & displacement)

displacement vector(dr) is of line joining initial & final position vectors...

W = -k(xi+yj).(dxi+dyj)

W = -2k[xy] lim (0,0) to (a,a)

W = -2ka2

this is the amount of work

10 years ago

there is a mistake in previous sol so dont be confused n go for this result ...

W = -K(yi+xj) . (dxi+dyj)

dW = -k(ydx + xdy)

dW = -kd(xy)

integrating both sides ,

W = -k [xy] lim from (0,0) to (a,a)

= -ka2

this is correct ....

10 years ago

thnku Mr.Vikas  Yash Chourasiya
askIITians Faculty 256 Points
one year ago
Hello Student

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

I hope this solution will help you.

## ASK QUESTION

Get your questions answered by the expert for free