Question #e4198

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Question #e4198

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Answer 1 · 2017-03-29T15:55:07

We know that the conservative force $\to F$ $vecF$ is related to the potential energy function U(x,y) in two dimension through the following relationship

$\to F = - \frac{\partial U}{\partial x} ˆ i - \frac{\partial U}{\partial y} ˆ j$ $vecF=-(delU)/(delx)hati-(delU)/(dely)hatj$

Here it is given that $U = 6 x + 6 y$ $U=6x+6y$

So

$\to F = - \frac{\partial (6 x + 8 y)}{\partial x} ˆ i - \frac{\partial (6 x + 8 y)}{\partial y} ˆ j$ $vecF=-(del(6x+8y))/(delx)hati-(del(6x+8y))/(dely)hatj$

$\Rightarrow \to F = (- 6 ˆ i - 8 ˆ j) N$ $=>vecF=(-6hati-8hatj)N$

So acceleration $\to a = \frac{\to F}{M}$ $veca=(vecF)/M$

where $M =$ $M=$ mass of the body moving $= 2 k g$ $=2kg$

Hence $\to a = \frac{\to F}{2} = (- 3 ˆ i - 4 ˆ j) m / s^{2}$ $veca=(vecF)/2=(-3hati-4hatj)m"/"s^2$

When the body crosses the Y-axis, then x=0.
Again it is given that body is at rest at the position (6m,4m) at t=0
This means
at x=6m, velocity =0;
When the body crosses Y-axis,its displacement along X-axis becomes $s_{x} = - 6 m$ $s_x=-6m$ .

If this displacement occurs in t sec due to its movement with acceleration along X axis $a_{x} = - 3 m / s^{2}$ $a_x=-3m"/"s^2$

then we can write

$s_{x} = \frac{1}{2} a_{x} t^{2}$ $s_x=1/2a_xt^2$

$\Rightarrow - 6 = \frac{1}{2} \times (- 3) t^{2}$ $=>-6=1/2xx(-3)t^2$

$\Rightarrow t = 2 s$ $=>t=2s$

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Question #e4198

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