Question

If a force f = (2x + 3x2 )î n acts along x-axis on an object and moves it from x = 2î m to x = 4î m, the work done is

Answer 1

Work done is the dot product of force vector and displacement vector
e.g., W = F.S

Here given ,
F = (2x + 3x²)i , here we see F is variable so, we have to use integration
So, work done , W = $\int\limits^{x_2}_{x_1}{F(x)},dx$
W = $\int\limits^{4}_{2}{(2x + 3x^2)},dx$
= $2\int\limits^{4}_{2}{x}\,dx+3\int\limts^{4}_{2}{x^2}\,dx$
= $[x^2]^4_2+[x^3]^4_2$
= (16 - 4) + (64 - 8)
= 12 + 56
= 68 J

Hence, answer is 68J

Answer 2

Answer:

68 J.

Explanation:

Work done on an object is given by

$W=\vec F \cdot \vec r$.

where,

$\vec F$ = force acting on the object.

$\vec r$ = displacement of the object.

If the force is variable, such that it is varying with the distance, then the net work done is given by

$W = \int \vec F \cdot d\vec r$.

$d\vec r = dx\ \hat i+dy\ \hat j+dz\ \hat k.$

$\hat i,\ \hat j,\ \hat k$ = unit vectors along x, y and z axes respectively.

Given:

$\vec F = (2x+3x^2)\hat i\ N$.

Therefore, the work done in moving the object from $x=2\hat i\ m$ to $x=4\hat i\ m$ is given by

$W=\int^{x=4\hat i}_{x=2\hat i}(2x+3x^2)\hat i\cdot(dx\hat i+dy\hat j+dz\hat k)\\=\int^{x=4\hat i}_{x=2\hat i}(2x+3x^2)dx\\=\left( 2\dfrac{x^2}{2}+3\dfrac{x^3}{3}\right)^{x=4}_{x=2}\\=\left( x^2+x^3\right)^{x=4}_{x=2}\\=(4^2+4^3)-(2^2+2^3)\\=80-12\\=68\ J.$