Question

By applying a force F=(3xy−5z)j+4zk a particle is moved along the path $y={{x}^{2}}$ from point (0,0,0) to the point (2,4,0). The work done by the F on the particle is (all values are in SI units)
$\begin{aligned} & a)\dfrac{280}{5}J \\\ & b)\dfrac{140}{5}J \\\ & c)\dfrac{232}{5}J \\\ & d)\dfrac{192}{5}J \\\ \end{aligned}$

Answer 1

In the above question, we can observe that the force is not constant. Earlier, we discussed that the work done of a force that changes as the particle or object travels must include integration. We need to write down the force in a single variable and integrate it.

Complete step by step answer:
Let us write down the given values and equations below,
$F=(3xy-5z)j+4zk$
As the equation of motion is given, let substitute it in the force equation and write it down in single variable,
$\begin{aligned} & y={{x}^{2}} \\\ & \Rightarrow x={{y}^{\dfrac{1}{2}}} \\\ & \therefore F=(3{{y}^{\dfrac{3}{2}}}-5z)j-4zk \\\ \end{aligned}$
If we solve the work done in y axis or j direction,
The work done will be,
$\begin{aligned} & dw=F.dx \\\ & \Rightarrow W=\int\limits_{0}^{4}{3{{y}^{\dfrac{3}{2}}}dy} \\\ & \Rightarrow W=\dfrac{6}{5}\times {{[{{y}^{\dfrac{5}{2}}}]}_{0}}^{4} \\\ & \therefore W=\dfrac{192}{5}J \\\ \end{aligned}$
We didn’t consider the z axis because if we observe the initial position and final position of the particle, the z axis did not change, it is just zero in both cases.
Hence, the work done is calculated in the above way.
The correct option is option d.

Additional information:
A force is said to do work when it acts on a body so that there is a displacement of the point of application in the direction of the force. Thus, a force does work when it results in movement. In the case of a variable force, integration is necessary to calculate the work done. For example, let’s consider work done by a spring. According to Hooke's law the restoring force (or spring force) of a perfectly elastic spring is proportional to its extension (or compression), but opposite to the direction of extension (or compression). The same integration approach can be also applied to the work done by a constant force. This suggests that integrating the product of force and distance is the general way of determining the work done by a force on a moving body.

Note:
In the above question, the force is not constant. As the object or particle moves from one place to another, the force changes thoroughly. So, we need to integrate the force applied with respect to the distance travelled to find out the work done.