Question

The force F = (4.99 N)i + (7.97 N)j acts on a particle as it moves along a straight line from Cartesian coordinate (2.66 m, 3.12 m) to (5.24 m, 8.49 m). what is the work done by the force?

Answer 1

Given data

*The given force is F = (4.99 N)i + (7.97 N)j

*The given initial position of the particle is r_i = (2.66i + 3.12j)

*The given final position of the particle is r_f = (5.24i + 8.49j)

The formula for change in the displacement of the particle is given as

`\Delta r=r_f-r_i`

Substitute the known values in the above expression as

`\begin{gathered} \Delta r=(5.24i+8.49j)-(2.66i+3.12j) \\ =(5.24i-2.66i)+(8.49j-3.12j) \\ =2.58i+5.37j \end{gathered}`

The formula for the work done by the force is given as

`W=F\text{.}\Delta r`

Substitute the known values in the above expression as

`\begin{gathered} W=(4.99i+7.97j).(2.58i+5.37j) \\ =55.67\text{ J} \end{gathered}`

Hence, the work done by the force is W = 55.67 J