The force on a particle is directed along an x axis and given by f = f0(x/x0 - 1) where x is in meters and f is in newtons. if f0 = 0.98 n and x0 = 4.9 m, find the work done by the force in moving the particle from x = 0 to x = 2x0 m.

W = 9.604 J

Explanation:

Given data:

Function of force with respect to displacement (x) as:

F = F₀(x/x₀ - 1)

F₀ = 0.98 N

x₀ = 4.9 m

on substituting the values in the given function, we get

F = 0.98 × [(x/4.9) - 1]

Now, the work done is given as:

W = F . dx

substituting the value of force, we have

W = 0.98 × [(x/4.9) - 1] . dx

on integerating the above formula for the limit x = 0 to x = 2x₀ = 2 × 4.9 = 9.8 m

we get

W =  

or

W =

or

W = 0.98 × [(9.8²/4.9 - 9.8) - 0]

or

W = 9.604 J

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Explanation:

I hope this helps! (This is in my own words, by the way) :>