Question

The force on a particle as the function of displacement x ( in x- direction ) is given by
F = 10 + 0.5x The work done corresponding to displacement of particle from x =0 to x = 2 unit is ​

Answer 1

⤴️ Answer in attachment

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Answer 2

A force f = 0.5x +10 acts on a particle. here f is in newton and x is in metre. we have to calculate the work done by the force during the displacement of the particle from x=0 to x=2 m.

we know, workdone is the dot product of force and displacement.

so, W = F.s

but when force is a function of x.i.e., f(x)

here, f = 0.5x + 10 and limits ; x = 0 to x = 2

so, W = $\int\limits^2_0 {(0.5+10)} \, dx$

= 0.5/2 (2² - 0²) + 10(2 - 0)

= 0.25 (4) + 20

= 21 J

hence, workdone by the force is 21J