Question

A particle is moving in the x-y plane, following the path y(x)= A[1 + sin(kx)]. It moves from x = 0 to x = c while it is subjected to the force F(x) = sin(kx) that is parallel to the y axis. Here A, k, Band care constants. * 50% Part (a) Write an expression for the work done by the force. WE(AB)4(1 - cos(k) (2)) X Attempts Remain 50% Part (

Answer 1

The work done by the force F(x) = sin(kx) parallel to the y-axis is calculated using the formula W = ∫F dx. The work done is given by (1/k)(1 - cos(k*c)).

Step-by-step explanation:

The work done by a force is given by the formula:

W = ∫F dx

We know that the force F(x) = sin(kx) is parallel to the y-axis, which means it only has a y-component. Therefore, the work done by this force can be calculated as:

W = ∫F dx = ∫sin(kx) dx = -cos(kx)/k

Now, we need to find the limits of integration. The particle moves from x = 0 to x = c, so the limits of integration are 0 and c:

W = ∫0csin(kx) dx = [-cos(kx)/k]0c = (1/k)(cos(k*0) - cos(k*c)) = (1/k)(1 - cos(k*c))

Answer 2

The work done by the force F(x) = sin(kx) parallel to the y-axis as a particle moves in the x direction from x = 0 to x = c is zero, because the force has no component in the direction of movement.

Step-by-step explanation:

To calculate the work done by the force F(x) = sin(kx) that is parallel to the y-axis as a particle moves from x = 0 to x = c in the x-y plane, you would use the work formula W = ∫ F dx, where F is the force and dx is the differential distance element in the direction of the force. Since the force is only in the y direction, and displacement dx is in the x direction, the angle between them is 90 degrees, resulting in zero work being done by this force in the x direction. If you need to find the work done along a different path, you would need to account for the component of the displacement in the direction of the force.