Question

The particle moves first along the y axis from the origin to the point (0,6m) and then parallel to the x axis until it reaches (3m,6m). it is subject to a force f⃗ =cxyi^+dj^, where c = 8.0 n/m2 and d = 13 n . calculate the work done by the force. express your answer using two significant figures.

Answer 1

First we calculate the force.

F = 8 *3 * 6 i + 13 j = 144 i + 13 j

F = sqrt (144^2 + 13^2) = 144.59 N

 

Then calculate the displacement d.

d = sqrt (3^2 + 6^2) = 6.71 m

 

Work is the produce of force and displacement, therefore:

W = F d = 144.59 N * 6.71 m

W = 970 J

Answer 2

The work done by the force on the particle which moves first along the y-axis from the origin and then parallel to the x-axis is 970 J.

How to find work done?

Work done is the force applied on a body to move it over a distance. Work done can be calculated with the following formula.

$W=F\times d$

Here (F) is the magnitude of force and (d) is the distance traveled.

The particle moves first along the y axis from the origin to the point (0,6m) and then parallel to the x axis until it reaches (3m,6m).

The particle is subject to a force,

$\vec F=cxy\hat i+d\hat j$

Here,  c = 8.0 N/m2 and d = 13 N. The final point of the displacement of the particle is (3 m, 6 m).

Put the values of (x,y), c and d to find the value of force.

$\vec F=(8)(3)(6)\hat i+(13)\hat j\\\vec F=144\hat i+13\hat j$

Solving this equation as,

$F=\sqrt{(144)^2+(13)^2}\\F=144.59\rm\; N$

The initial position of the particle is at the origin (0, 0) and the final position of the particle is (3 m, 6 m). Thus, the displacement is,

$d=\sqrt{(3-0)^2+(6-0)^2}\\d=6.71\rm\; m$

The work done will be,

$W=144.59\times(6.71)\\W=970\rm\; J$

Hence, the work done by the force on the particle which moves first along the y-axis from the origin and then parallel to the x-axis is 970 J.

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