Question #137918

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).

Expert's answer

A=∣F⃗∣⋅∣s⃗∣⋅cos(φ)=F⃗⋅s⃗A=|\vec F|\cdot |\vec s|\cdot cos(φ)=\vec F\cdot \vec s ,

where ∣F⃗∣|\vec F| (N, Newton) is the vector F⃗\vec F length, ∣s⃗∣|\vec s| (m, meter) is the vector s⃗\vec s length

( s⃗=r⃗2−r⃗1=(2i⃗−3j⃗)−(i⃗+2j⃗)=i⃗−5j⃗\vec s=\vec r_{2}-\vec r_{1}=(2\vec i-3\vec j)-(\vec i+2\vec j)=\vec i-5\vec j ), φφ is the angle between F⃗\vec F and s⃗\vec s .

Thereby ∣F⃗∣⋅cos(φ)|\vec F|\cdot cos(φ) is the length of the projection of the vector F⃗\vec F to the vector s⃗\vec s .

A=F⃗⋅s⃗=(3i⃗+4j⃗)(i⃗−5j⃗)=A=\vec F\cdot \vec s=(3\vec i+4\vec j)(\vec i-5\vec j)=

=3i⃗2−15i⃗j⃗+4j⃗i⃗−20j⃗2=3−0+0−20==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=3-20=-17J (Joule).

Answer: −17J.


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