An object is displaced from position vector $r_{1}$ =(2i+3j)m to $r_{2}$ =(4i+6j)m under the action of a force F=(3 $x^{2}$ i+2yj)N. Finf the work done by this force.

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Solution

Given,

$r_{1} = (2 i + 3 j) m$

$r_{2} = (4 i + 6 j) m$

$F o r c e = (3 x^{2} i + 2 y j) N$

So, 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 $6 (4 + 8 t + 4 t^{2}) + 6 (3 + 3 t)$ which is $42 + 66 t + 24 t^{2}$ from $0 t o 1$ ,

After that we get,

$42 + 33 + 8 = 83 J$

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Similar questions

{\to r}_{1} = (2^i + 3^j) m

{\to r}_{2} = (4^i + 6^j) m

\to F = (3 x^{2}^i + 2 y^j) N

r_{1} = (2^i + 3^j) m

r_{2} = (4^i + 6^j) m

F = (3 x^{2}^i + 2 y^j) N

{\to r}_{1} =^i + 2^j m

_{2} = 2^i + 4^j m

\to F = (6 x^{2}^i + 4 y^j) N

{\to r}_{1} = (2^i + 3^j) m

{\to r}_{2} = (4^i + 6^j) m

\to F = (3 x^{2}^i + 2 y^j) N

_{1} = (2^i + 3^j) m

_{2} = (4^i + 6^j) m

F = (3 x^{2}^i + 2 y^j) N

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