Question

An object is displaced from position vector r1 = (2i + 3j) m to r2 = (4i + 6j)m under a force vector f =(3x^2i+2yj) n. find the work done by this force.

Answer 1

Answer

Ok, the path that the object takes is defined by... 

(2,3)+t(2,3) where 0<t<1. 

Thus, the work done is equal to the integral of the dot product between force and the derivative of position with respect to time from 0 to 1. 

This is the integral of 6(4+8t+4t^2)+6(3+3t) which is 42+66t+24t^2 from 0 to 1, which comes out to be... 

42+33+8 which is... 

83 J!!!

Answer 2

Answer:

The work done by  force is 83 J.

Explanation:

Given:

• Position vectors,

$\mathrm{r}_{1}=(2 \mathrm{i}+3 \mathrm{j}) \mathrm{m}\\\mathrm{r}_{2}=(4 \mathrm{i}+6 \mathrm{j}) \mathrm{m}$

• Force $=\left(3 x^{2} i+2 y j\right) N$

So, the path that the object takes is defined by,

$(2,3)+t(2,3) where 0 < t < 1$

The work done is equal to the integral of the dot product between force and the derivative of position with respect to time from o to 1.

$Work done is given by, W=\int \vec{F} \cdot(d x \hat{i}+d y \hat{j}) W=\int_{2}^{4} 3 x^{2} d x+\int_{3}^{6} 2 y d y W=\left[x^{3}\right]_{2}^{4}+\left[y^{2}\right]_{3}^{6}$

                                             $W=4^{3} - 2^{3}+ 6^{2}- 3^{2} \\W = (64-8) + (36-9)\\W = 56 + 27\\W = 83 J$

Work done = 83J