Respuesta :
Answer:
W = 9.604 J
Explanation:
Given data:
Function of force with respect to displacement (x) as:
F = Fâ‚€(x/xâ‚€ - 1)
Fâ‚€ = 0.98 N
xâ‚€ = 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 = Â [tex]\int\limits^{9.8}_00.98\times[(x/4.9) - 1]} \, dx[/tex]
or
W = [tex]0.98\times[(x^2/4.9 - x)]^{9.8}_0[/tex]
or
W = 0.98 × [(9.8²/4.9 - 9.8) - 0]
or
W = 9.604 J
Work done by the force on a particle is the amount of energy required to displace it. The total work done  by the force in moving the particle from 0 to [tex]2x_0[/tex] position is zero.
What is work done by the force?
The amount of energy which is required to change the position of a particle from one place to another by the application of force, is called the work done by the force.
It can be given as,
[tex]W=Fdx[/tex]
Here, (F) is the force applied and (x) is the distance of the body.
Given information-
The equation of the force is given in the problem is,
[tex]F = F_0(\dfrac{x}{x_0} - 1)[/tex]
The force on the partial is 0.98 sec.
The distance of x axis is 4.9 meters.
Put the value of force in the formula of work done as,
[tex]W = F_0(\dfrac{x}{x_0} - 1)dx[/tex]
Value of  [tex]2x_0[/tex]  is,
[tex]2x_0=2\times4.9\\2x_0=9.8[/tex]
Thus, the particle moves from 0 to 9.8 meters.
Therefore put the values in the above equation by integrating it with limits 0 to 9.8 as,
[tex]W=0.98\int\limits^{9.8}_0({\dfrac{x}{9.8}-1}) \, dx\\W=0.98[{\dfrac{x^2}{2\times4.9}-x}]^{9.8}_0 \\W=0.98[{\dfrac{(9.8)^2}{2\times4.9}-9.8-0+0}] \\W=0[/tex]
Hence, the total work done  by the force in moving the particle from 0 to [tex]2x_0[/tex] position is zero.
Learn more about the work done by force here;
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