Question

A position dependent force F = 7-2X+3X^2N acts on a small body of mass 2kg and displaces it from=0 to X=5m.the work done in joule is.

Answer 1

Answer: The work done is 135 J.

Explanation:

Given that,

Force $F = 7-2x+3x^{2} N$

Mass m = 2 kg

Position x = 0 to x = 5m

We know that,

The work done is

$W = \int_{x_{1}}^{x_{2}}Fdx$

$W = \int_{0}^{5}(7-2x+3x^{2})dx$

$W =(7x-x^{2}+x^{3})_{0}^{5}$

$W= (35-25+125)J$

$W = 135 J$

Hence, the work done is 135 J.

Answer 2

"Given data states that the F = 7 - 2x + 3x 2 acting on a small body with 2 kg mass which is being displaced from x = 0 to 5 m. therefore, differentiating work with respect to displacement gives force, hereby we can integrate the said equation and get the value of work done.  

$F\quad =\quad \frac { dw }{ dx }$

$\Rightarrow Fdx\quad =\quad dw$

$\Rightarrow \int { Fdx } \quad =\quad \int { dx }$

$\Rightarrow W = \int _{ 0 }^{ 5 }{ 7-2x+3{ x }^{ 2 } } d x$

$\Rightarrow W = { \left[ 7x-\frac {2({ x }^{ 2 }) }{ 2 } +\frac {3({ x }^{ 3 }) }{ 3 } \right] }_{ 0 }^{ 5 }$

$\Rightarrow w = [ 35 - 25 + 125 ]$

$\Rightarrow w = 135 Joules$"