Question

A force of 2i+3j units has its point of application moved from point A (1,3) to point B(5,7) then the work of force is​

Answer 1

Answer:

20 joules..............

Answer 2

Answer:

What is the work done by force F=2i^-3j^+k^ when its points of application move the point (1,2,-3) to the point (2,0,-5)?

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Work done by force acting on a object/body can be described as formula below

W=F∗s

W= work done

F= force acting

s= displacement of the object/body

So work is force multiplied by displacement

But when the force and displacement or given in vector form we use dot product or scalar product as the work done is a scalar , like below

W=F⃗ ⋅s⃗

Here the force the force is already give as F⃗ =2i^−3j^+k^

And we are given that it moves from point (1,2,-3) to the point (2,0,-5)

The displacement vector s with initial point (x1, y1, z1) and terminal point (x2, y2, z2) is s = (x2 − x1, y2 − y1, z2 − z1).

s⃗ =⟨2−1,0−2,(−5−(−3))⟩=⟨1,−2,−2⟩or

s⃗ =i^−2j^−2k^

So now work done is W=F⃗ ⋅s⃗ =⟨2i^−3j^+k^⟩⋅⟨i^−2j^−2k^⟩=⟨2∗1+(−3)∗(−2)+1∗(−2)⟩

W=6

So work done is 6 Joules (assuming force is in newtons and displacement in meters)