A particle experiences a variable for $\vec{F} = (4 x \hat{i} + 3 y^{2} \hat{j})$ in a horizontal x-y plane. Assume distance in meters and force is Newton. if the particle moves from point (1, 2) to point (2, 3) in the x-y plane, then kinetic Energy changes by :

Option 1 -

50.0 J

Option 2 -

12.5 J

Option 3 -

25.0 J

Option 4 -

0 J

2 Views | Posted 2 months ago

Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

2 months ago

Correct Option - 3

Detailed Solution:

$\vec{F} = 4 x \hat{i} + 3 y^{2} \hat{j}$

$\begin{array}{l} d \vec{r} = d \vec{s} = d x \hat{i} + d y \hat{j} \\ \int d w = \int \vec{F} . d \vec{r} \\ = \int (4 x \hat{i} + 3 y^{2} \hat{j}) . (d x \hat{i} + d y \hat{j}) \\ = \int_{1}^{2} 4 x d x + \int_{2}^{3} 3 y^{2} d y \\ = \frac{4}{2} {[x^{2}]}_{1}^{2} + \frac{3}{3} {[y^{3}]}_{2}^{3} \\ = 2 [4 - 1] + [27 - 8] \end{array}$

= 6 + 19 = 25 J

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A particle experiences a variable for $\vec{F} = (4 x \hat{i} + 3 y^{2} \hat{j})$ in a horizontal x-y plane. Assume distance in meters and force is Newton. if the particle moves from point (1, 2) to point (2, 3) in the x-y plane, then kinetic Energy changes by :

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A particle experiences a variable for F → = ( 4 x i ^ + 3 y 2 j ^ ) in a horizontal x-y plane. Assume distance in meters and force is Newton. if the particle moves from point (1, 2) to point (2, 3) in the x-y plane, then kinetic Energy changes by :

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A particle experiences a variable for $\vec{F} = (4 x \hat{i} + 3 y^{2} \hat{j})$ in a horizontal x-y plane. Assume distance in meters and force is Newton. if the particle moves from point (1, 2) to point (2, 3) in the x-y plane, then kinetic Energy changes by :