1) A particle moves in the xy plane (see the from the origin to a point having coordinates x = 7m and y = 5m under the influence of a force given by F= 3yi^2 - 5xj.
a) What is the work done on the particle by the force F if it moves along path 1? (ABD) J
b) What is the work done on the particle by the force F if it moves along path 2? (ACD) J
c) What is the work done on the particle by the force F if it moves along path 3? (AD) J
The plane is a square
C--------D
l0000000l
l0000000l
A--------B
To find the work done by a force, we need to integrate the dot product of the force vector and the displacement vector along the path of the particle.
a) For path 1 (ABD), we need to find the work done along the path from A to B, and then from B to D.
- From A to B, the force F is given by F = 3yi^2 - 5xj. The displacement vector is given by AB = 7i.
- The work done from A to B is calculated by integrating the dot product of F and AB: ∫(F · AB) = ∫(3y + 35x)dx, from x = 0 to x = 7.
- Evaluating the integral, we get ∫(3y + 35x)dx = ∫(3(5) + 35x)dx = 15x + 17.5x^2 | from x = 0 to x = 7.
- Plugging in the values, we have (15 * 7 + 17.5 * 7^2) - (15 * 0 + 17.5 * 0^2) = 105 + 17.5 * 49 - 0 = 105 + 857.5 = 962.5 J.
- From B to D, the force F is given by F = 3yi^2 - 5xj. The displacement vector is given by BD = 5j.
- The work done from B to D is calculated by integrating the dot product of F and BD: ∫(F · BD) = ∫(3y + 35x)dy, from y = 5 to y = 0.
- Evaluating the integral, we have ∫(3y + 35x)dy = 3y^2/2 + 35xy | from y = 5 to y = 0.
- Plugging in the values, we get (3(0)^2/2 + 35x(0)) - (3(5)^2/2 + 35x(5)) = 0 - (75/2 + 175x) = -75/2 - 175x J.
The total work done along path 1 is the sum of the work done from A to B and from B to D: 962.5 J + (-75/2 - 175x) J.
b) For path 2 (ACD), we need to find the work done along the path from A to C and from C to D.
- From A to C, the force F is given by F = 3yi^2 - 5xj. The displacement vector is given by AC = 5i.
- The work done from A to C is calculated by integrating the dot product of F and AC: ∫(F · AC) = ∫(3y + 35x)dx, from x = 0 to x = 5.
- Evaluating the integral, we get ∫(3y + 35x)dx = 15x + 17.5x^2 | from x = 0 to x = 5.
- Plugging in the values, we have (15 * 5 + 17.5 * 5^2) - (15 * 0 + 17.5 * 0^2) = 75 + 17.5 * 25 - 0 = 75 + 437.5 = 512.5 J.
- From C to D, the force F is given by F = 3yi^2 - 5xj. The displacement vector is given by CD = -7i.
- The work done from C to D is calculated by integrating the dot product of F and CD: ∫(F · CD) = ∫(3y + 35x)dx, from x = 5 to x = 0.
- Evaluating the integral, we get ∫(3y + 35x)dx = 15x + 17.5x^2 | from x = 5 to x = 0.
- Plugging in the values, we have (15 * 0 + 17.5 * 0^2) - (15 * 5 + 17.5 * 5^2) = 0 - (75 + 17.5 * 25) = -75 - 437.5 = -512.5 J.
The total work done along path 2 is the sum of the work done from A to C and from C to D: 512.5 J + (-512.5) J = 0 J.
c) For path 3 (AD), we need to find the work done along the direct path from A to D.
- The force F is given by F = 3yi^2 - 5xj. The displacement vector is given by AD = 7i - 5j.
- The work done from A to D is calculated by integrating the dot product of F and AD: ∫(F · AD) = ∫(3y + 35x)dt, from t = 0 to t = 1.
- Since AD can be expressed as AD = 7i - 5j = (7i - 0j) - (0i - 5j), we can calculate the work done as the sum of the work done along the x-direction and the work done along the y-direction.
- The work done along the x-direction is ∫(35x)dt = 35xt | from t = 0 to t = 1, which is equal to 35 * 7 = 245 J.
- The work done along the y-direction is ∫(3y)dt = 3yt | from t = 0 to t = 1, which is equal to 3 * (-5) = -15 J.
The total work done along path 3 is the sum of the work done along the x-direction and the work done along the y-direction: 245 J + (-15) J = 230 J.
Therefore, the answers are:
a) The work done on the particle by the force F along path 1 (ABD) is 962.5 J.
b) The work done on the particle by the force F along path 2 (ACD) is 0 J.
c) The work done on the particle by the force F along path 3 (AD) is 230 J.