The work done by a force is given by the line integral of the force along the path of the displacement. In this case, the force is given by F = (4y, x^2) and the path is x = y from (0,0) to (4,4).

The work done, W, is given by:

W = ∫ F . dr

where F is the force, dr is the differential displacement vector, and "." denotes the dot product.

Since the path is along x = y, we can write dr = (dx, dx) and F = (4x, x^2).

So,

W = ∫ from 0 to 4 (4x, x^2) . (dx, dx)
= ∫ from 0 to 4 (4x*dx + x^2*dx)
= ∫ from 0 to 4 4x*dx + ∫ from 0 to 4 x^2*dx
= [2x^2] from 0 to 4 + [x^3/3] from 0 to 4
= 2*(4^2) - 2*(0^2) + (4^3)/3 - (0^3)/3
= 32 + 64/3
= 96/3 + 64/3
= 160/3 J

So, the work done by the force on the body is 160/3 Joules.

Question

The work done by a force is given by the line integral of the force along the path of the displacement. In this case, the force is given by F = (4y, x^2) and the path is x = y from (0,0) to (4,4).

The work done, W, is given by:

W = ∫ F . dr

where F is the force, dr is the differential displacement vector, and "." denotes the dot product.

Since the path is along x = y, we can write dr = (dx, dx) and F = (4x, x^2).

So,

W = ∫ from 0 to 4 (4x, x^2) . (dx, dx)
  = ∫ from 0 to 4 (4x*dx + x^2*dx)
  = ∫ from 0 to 4 4x*dx + ∫ from 0 to 4 x^2*dx
  = [2x^2] from 0 to 4 + [x^3/3] from 0 to 4
  = 2*(4^2) - 2*(0^2) + (4^3)/3 - (0^3)/3
  = 32 + 64/3
  = 96/3 + 64/3
  = 160/3 J

So, the work done by the force on the body is 160/3 Joules.

Knowee AI · Accepted Answer

The work done by a force is given by the line integral of the force along the path of the displacement. In this case, the force is given by F = (4y, x^2) and the path is x = y from (0,0) to (4,4).

The work done, W, is given by:

W = ∫ F . dr

where F is the force, dr is the differential displacement vector, and "." denotes the dot product.

Since the path is along x = y, we can write dr = (dx, dx) and F = (4x, x^2).

So,

W = ∫ from 0 to 4 (4x, x^2) . (dx, dx)
  = ∫ from 0 to 4 (4x*dx + x^2*dx)
  = ∫ from 0 to 4 4x*dx + ∫ from 0 to 4 x^2*dx
  = [2x^2] from 0 to 4 + [x^3/3] from 0 to 4
  = 2*(4^2) - 2*(0^2) + (4^3)/3 - (0^3)/3
  = 32 + 64/3
  = 96/3 + 64/3
  = 160/3 J

So, the work done by the force on the body is 160/3 Joules.

A body is displaced from (0, 0) to (4m, 4m) along thepath x = y by a force ~F = (4yˆi+x2ˆj) N.Estimate the work done by this force on the body.

Question

A body is displaced from (0, 0) to (4m, 4m) along the path x = y by a force $\vec{F} = (4y \hat{i} + x^2 \hat{j}) \, \text{N}$ . Estimate the work done by this force on the body.

Solution

Similar Questions

Upgrade your grade with Knowee