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How much work is done by a force F→=(2xN)i^+(3N)j^,with x in meters, that moves a particle from a position ri→=(2m)i^+(3m)j^to a position rf→=-(4m)i^+(3m)j^?

Short Answer

Expert verified

Work done by the applied force is -6J

Step by step solution

01

Given information

It is given that,

Fapp=(2xN)i^+(3N)j^ri→=(2m)i^+(3m)j^rf→=-(4m)i^-(3m)j^

02

Determining the concept

This problem deals with the variable forces. When the direction and the amount of force vary throughout the motion, variable forces occur. To solve this problem, usetheconcept of variable forces. Work done due to variable forces can be found by using integration.

Formula:

The work done is given by,

W=∫Fx.dx+∫Fy.dy

Where,

Fx,FyAre the forces in x and y direction, dx and dy are small displacements.

03

Determining the work done by the applied force (Wapp).

Using the above formula, writethetotal work done as,

W=∫xixfFxdx+∫yiyfFydy

Substituting the value of forces and from the given data,

W=∫2-42×dx+∫3-33dyW=x22-4+3y3-3W=16-4-9-9W=-6J

Work done by the applied force is -6J

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