Using parameterisation to calculate work done by force

The final answer should be a positive number, not a negative one.In summary, a force given by F = -K(yi + xj) acts on a particle moving in the x-y plane. The particle is taken from the origin to the point (a, 0) along the positive x-axis and then to the point (a, a) parallel to the y-axis. The total work done by the force F on the particle is -ak, which is incorrect. The correct answer should be a positive number.
  • #1
feetnappy
A force F = -K(yi + xj) (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 the point (a, 0) and then parallel to the y-axis to the point (a, a). What is the total work done by the force F on the particle.

My attempt:

As the final position of the particle is (a, a) so I get the following using parameterisation for 0 <= t <= 1

x = at
y = at

Thus the work done,

W = integration -k(ydx + xdy)
= integration -k(atdt + atdt)
= integration -k(2atdt)
= -2ak integration (tdt)
= -ak [t]^2
= -ak

What am I doing wrong here?
 
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  • #2
What are dx and dy?
 
  • #3
feetnappy said:
A force F = -K(yi + xj) (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 the point (a, 0) and then parallel to the y-axis to the point (a, a). What is the total work done by the force F on the particle.

My attempt:

As the final position of the particle is (a, a) so I get the following using parameterisation for 0 <= t <= 1

x = at
y = at

Thus the work done,

W = integration -k(ydx + xdy)
= integration -k(atdt + atdt)
= integration -k(2atdt)
= -2ak integration (tdt)
= -ak [t]^2
= -ak

What am I doing wrong here?

Why do you go along a straight line from (0,0) to (a,a)? The question said to go along the sides of the square. Of course, for some types of forces the work done is independent of the path, but is that the case here?

Anyway, even for the diagonal path, have you double-checked your work?
 
Last edited:

FAQ: Using parameterisation to calculate work done by force

1. What is parameterisation and how is it used to calculate work done by force?

Parameterisation is a mathematical technique used to express a function or equation in terms of a parameter. In the context of calculating work done by force, parameterisation involves breaking down the force into its individual components and integrating them with respect to the displacement along a specific parameter, such as time or distance. This allows for a more accurate and efficient calculation of work done.

2. What are the benefits of using parameterisation to calculate work done by force?

Using parameterisation allows for a more precise and accurate calculation of work done by force, as it takes into account the direction and magnitude of the force at each point along the displacement. It also allows for a more efficient calculation, as it simplifies the integration process.

3. Can parameterisation be used for any type of force?

Yes, parameterisation can be used for any type of force, whether it is a constant force, a variable force, or a force that changes direction. As long as the force can be broken down into its components and integrated with respect to a specific parameter, parameterisation can be used to calculate the work done by that force.

4. Are there any limitations to using parameterisation for calculating work done by force?

While parameterisation is a useful technique for calculating work done by force, it does have its limitations. It may not be suitable for more complex forces or systems with multiple forces acting at once. In these cases, other methods such as numerical integration may be more appropriate.

5. How can I apply parameterisation to real-world scenarios?

Parameterisation can be applied to any real-world scenario where work is being done by a force. For example, it can be used to calculate the work done by a person lifting weights at the gym, or the work done by a car engine to move a vehicle. By breaking down the force into its components and integrating with respect to a specific parameter, you can accurately calculate the work done in these situations.

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