The potential energy (in joule) of a body of mass 2 kg moving in the x-y plane is given by
U = 6x + 8y
where the position coordinates x and y are measured in metre. If the body is at rest at point (6 m, 4m) at time t = 0, it will cross the y-axis at time t equal to
The potential energy (in joule) of a body of mass 2 kg moving in the x-y plane is given by
U = 6x + 8y
where the position coordinates x and y are measured in metre. If the body is at rest at point (6 m, 4m) at time t = 0, it will cross the y-axis at time t equal to
The correct option is B. 2 s.
Given that,
Potential energy = $U = 6x + 8y$
Now, \(F_x = \frac{-dU}{dx} = \frac{-d}{dx}(6x+8y)=-6~N\)
Therefore, the x component of acceleration is
\(a_x = \frac{F_x}{m} = \frac{-6}{2} = -3~ms^{-2}\)
The x coordinate of the body at time t is
\(x=x_0+\frac{1}{2}a_x t^2 = 6 - \frac{1}{2}\times 3\times t^2\)
\(x=\left ( 6-\frac{3}{2}t^2 \right )\)
The body will cross the y-axis when x = 0, i.e. at time t given by
\(\left ( 6-\frac{3}{2}t^2 \right ) = 0\) or $t = 2 s$.
Hence, the correct choice is (b).
The correct option is B. 2 s.
Given that,
Potential energy = $U = 6x + 8y$
Now, \(F_x = \frac{-dU}{dx} = \frac{-d}{dx}(6x+8y)=-6~N\)
Therefore, the x component of acceleration is
\(a_x = \frac{F_x}{m} = \frac{-6}{2} = -3~ms^{-2}\)
The x coordinate of the body at time t is
\(x=x_0+\frac{1}{2}a_x t^2 = 6 - \frac{1}{2}\times 3\times t^2\)
\(x=\left ( 6-\frac{3}{2}t^2 \right )\)
The body will cross the y-axis when x = 0, i.e. at time t given by
\(\left ( 6-\frac{3}{2}t^2 \right ) = 0\) or $t = 2 s$.
Hence, the correct choice is (b).
