Answer to Question #137952 in Mechanics | Relativity for Sneha

Question

A particle has shifted along some trajectory in the plane xy from point 1 whose radius vector r1 = i + 2j to point 2 with the radius vector r2 = 2i - 3j. During that time the particle experienced the action of certain forces, one of which being F = 3i + 4j. Find the work performed by the force F. (Here r1, r2, and F are given in SI units).

Answer 1

$A=|\vec F|\cdot |\vec s|\cdot cos(φ)=\vec F\cdot \vec s$ ,

where $|\vec F|$ (N, Newton) is the vector $\vec F$ length (or the value of force); $|\vec s|$ (m, meter) is the vector $\vec s$ length (or the length of the trajectory from point 1 to point 2)

($\vec s=\vec r_{2}-\vec r_{1}=(2\vec i-3\vec j)-(\vec i+2\vec j)=\vec i-5\vec j$ ), $φ$ is the angle between $\vec F$ and $\vec s$ .

Thereby $|\vec F|\cdot cos(φ)$ is the length of the projection of the vector $\vec F$ to the vector $\vec s$ .

$A=\vec F\cdot \vec s=(3\vec i+4\vec j)(\vec i-5\vec j)=$

$=3\vec i^2-15\vec i\vec j+4\vec j\vec i-20\vec j^2=3-0+0-20=$

$=3-20=-17J$ (Joule).

Answer: −17J.