Work & Vectors: Finding Force Work from Point to Point

[silentsaber]

Homework Statement

A force F = (2.7i + 3.2j) N is acting on a particle while it travels from the point (x, y) =
(1, -1) m to the point (2, 1) m. Find the work done by the force F on the particle.

Homework Equations

W=FD and A=SqRt of Ax^2+Ay^2

The Attempt at a Solution

ok i used the right triangle to find the resultant vector of the force F by squaring both the 2.7 and 3.2 then adding them and take the square root i got 4.18 so that should be the F in W=FD and then i foudn the reultant vector of the coordinates 1,-1 and 2,1 and got square root of 5. which should be the distance and then i multiplied 4.18 and radical 5 and got 9.34 which is wrong.. where did i go wrong? btw 9.10 is the answer

 

[alphysicist]

Hi silentsaber,

Homework Statement

A force F = (2.7i + 3.2j) N is acting on a particle while it travels from the point (x, y) =
(1, -1) m to the point (2, 1) m. Find the work done by the force F on the particle.

Homework Equations

W=FD and A=SqRt of Ax^2+Ay^2

The Attempt at a Solution

ok i used the right triangle to find the resultant vector of the force F by squaring both the 2.7 and 3.2 then adding them and take the square root i got 4.18 so that should be the F in W=FD and then i foudn the reultant vector of the coordinates 1,-1 and 2,1 and got square root of 5. which should be the distance and then i multiplied 4.18 and radical 5 and got 9.34 which is wrong.. where did i go wrong? btw 9.10 is the answer

The problem is that the formula for work is not $W=Fd$ (product of magnitudes of force and displacement), it is

$$W=\vec F\cdot\vec d$$

(dot product of force and displacement vectors). Are you familiar with the dot product? If so, then once you find the two vectors involved the problem is quite straightforward.

 

[phreak]

You're working with vectors. So you need to find the dot product. W = F dot D.

 

[silentsaber]

thankyou guys very much i totally forgot the dot product existed XD