Question

The force f=2xi+2j+3z2k n is acting on a particle. Find the work done by this force in displacing the body from (1,2,3)m to (3,6,1)m

Answer 1

Explanation:

It is given that,

The force acting on the particle, $F=2xi+2yj+3z^2k$

Initial position of the particle, $x_1=i+2j+3k$

Final position of the particle, $x_2=3i+6j+k$

Work done is given by :

$W=\int\limits {F.dx}$

dx is the position of the particle, $dx=xi+yj+zk$

$W=\int\limits {(2xi+2yj+3z^2k).(xi+yj+zk)}$

$W=\int\limits {2x^2+2y+3z^3}$

$W=\int\limits^3_1 {2x^2}+\int\limits^6_2 {2y}+\int\limits^1_3 {2z^3}$

$W=\dfrac{2x^3}{3}|_1^3+y^2|_2^6+\dfrac{z^4}{2}|_3^1$

$W=\dfrac{2}{3}[3^3-1]+[6^2-2^2]+[1^4-3^4]$

W = -30.66 Joules

So, the work done by this force in displacing the body from (1,2,3)m to (3,6,1)m is -30.66 Joules. Hence, this is the required solution.

Answer 2

hlo mate here is ur ans.......

F = 2xi^+2j^​+3z2k

⇒Fx​ ​= 2x   Fy​ ​= 2    F2​ = 3z2

Work done, W=∫F.dr

⇒W=∫F.dr=∫13​2x.dx+∫26​2y.dy+∫31​3z2.dx

=[x2]13​+[y2]26​+[z3]1/3

  =8+32−26

 =14J.

Hence, the answer is 14J.

hoope this helps  uh dear :)

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