Question

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.

Answer 1

Answer:

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 =  $\int\limits^{9.8}_00.98\times[(x/4.9) - 1]} \, dx$

or

W = $0.98\times[(x^2/4.9 - x)]^{9.8}_0$

or

W = 0.98 × [(9.8²/4.9 - 9.8) - 0]

or

W = 9.604 J