Question

The force F = (3.77 N)i + (7.73 N)j acts on a particle as it moves first along the x axis from x = 1.81 m to x = 6.43 m, then parallel to the y axis from y = 4.97 m to y = 9.47 m. what is the work done by the force.

Answer 1

Given data:

* The force acting on the particle is,

`F=3.77\text{ i + 7.73 j}`

* The displacement of the particle along the x-axis is,

`\begin{gathered} dx=6.43-1.81 \\ dx=4.62\text{ m} \end{gathered}`

* The displacement of the particle along the y-axis is,

`\begin{gathered} dy=9.47-4.97 \\ dy=4.5\text{ m} \end{gathered}`

Solution:

The displacement vector of the particle is,

`\begin{gathered} d=\text{dx i + dy j} \\ d=4.62\text{ i + }4.5\text{ j} \end{gathered}`

The work done on the particle in terms of the force and displacement vector is,

`W=F\cdot d`

Substituting the known values,

`\begin{gathered} W=(3.77i+7.73j)\cdot(4.62i+4.5j) \\ W=17.42+34.78 \\ W=52.2\text{ J} \end{gathered}`

Thus, the work done by the force on the particle is 52.2 J.