Question

A force f=6x^2-4x-8 displaces a body from x=1m to x=2m work done is

Answer 1

Answer:

0 J

Explanation:

Given :

F = 6 x² - 4 x - 8

Displacement = 1 to 2 m

We know :

$\displaystyle{\text{W}=\int\limits^{x_1}_{x_2} {\text{F}} \, dx }$

$\displaystyle{\text{W}=\int\limits^{2}_{1} {(6x^2-4x-8)} \, dx }$

$\displaystyle{\text{W}=\left|\frac{6x^3}{3}-\frac{4x^2}{2}-\frac{8x}{1} \right|^2_1 }$

$\displaystyle{\text{W}=\left|2x^3-2x^2-8x|^2_1}$

W = ( 2.2³ - 2.2² - 8.2 ) - ( 2.1³ - 2.1² - 8.1 )

W = ( 16 - 8 - 16 ) - ( 2 - 2 - 8 )

W = - 8 + 8 J

W = 0 J

Therefore work done by body is 0 J

Answer 2

Answer:

The work done here is 0.

Explanation:

For work done by a variable force we have,

 $W = \int\limits^a_b {Force} \, dx$

For this question we have,

  $W= \int\limits^2_1 (6x^{2} {-4x-8)} \, dx$

Which on further solving gives,

$\displaystyle{\text{W}=\left\frac{6x^3}{3}-\frac{4x^2}{2}-\frac{8x}{1} \right|^2_1 }$

Which equals,

$W = [ \frac{6 (2^{3})}{3} -\frac{4(2^{2}) }{2} -8(2)]- [ \frac{6 (1^{3})}{3} -\frac{4(1^{2}) }{2} -8(1)]\\\\ \\ = (16-8-16)-(2-2-8) \\\\\\\\ = (-8)-(-8) \\\\ = 0$