Question

the work done by a force F=2(x+4y)i+8xj on a particle moving from origin to (4,2,0) along the path x2=8y is 10n j.Find n

Answer 1

Answer:

n=8

Explanation:

F = 2(x+4y)j^+8xj^

w=∫F¯.d¯r

= ∫(2x+8y)i^+(8xj^).(dxi^+dyj^)

= ∫(2x+8y)dx+8xdy=∫2xdx+8∫d(xy)

= [2x22]0   04   2+8(xy)

=[8y]02+8[xy]4,2

16+8[4×2−0]

= 16+64

=80N

n=8

Answer 2

Answer:

Value of n is 8.

Explanation:

Given force applied on a particle, $F=2(x+4y)\hat i+8x\hat j$

Particle is moved from O(0, 0, 0)  from origin to a point P(4,2,0)

Path followed is $x^{2} =8y$                                                        ...(1)

Differentiating the above equation,

                   $2xdx=8dy\implies xdx=4dy\implies dy=\frac{x}{4} dx$       ...(2)

Work done is given by $W=\int F.rdr$    

                                          $=\int (2(x+4y)\hat i+8x\hat j).(dx\hat i+dy\hat j)$

Substituting equations (1) and (2)

Work done,  $W=\int (2(x+4\frac{x^{2} }{8})\hat i+8x\hat j).(dx\hat i+\frac{x}{4} dx\hat j)$

                          $=\int ((2x+{x^{2} }})\hat i+8x\hat j).(dx\hat i+\frac{x}{4} dx\hat j)$

                          $=\int ((2x+{x^{2} }})\hat i.dx\hat i+8x\hat j.\frac{x}{4} dx\hat j$

                          $=\int (2x+{x^{2} }).dx+2x^{2} dx$

                          $=\int (2x+3{x^{2} })dx$

Integrating and using the limits for x,

                $W=\int\limits^4_0 (2x+3{x^{2} })dx=[\frac{2x^{2} }{2} +\frac{3x^{3} }{3} ]_{0}^{4}$

                     $=16 +64=80J$

Given work done in moving the particle, $W=10nJ$

Therefore, $80=10n \implies n=8$