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).
1
Expert's answer
2020-10-15T10:41:25-0400

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

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

( s=r2r1=(2i3j)(i+2j)=i5j\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 F\vec F and s\vec s .

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

A=Fs=(3i+4j)(i5j)=A=\vec F\cdot \vec s=(3\vec i+4\vec j)(\vec i-5\vec j)=

=3i215ij+4ji20j2=30+020==3\vec i^2-15\vec i\vec j+4\vec j\vec i-20\vec j^2=3-0+0-20=

=320=17J=3-20=-17J (Joule).

Answer: −17J.


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