A ball of mass 2 kg is moving in xy-plane wi | vedclass

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Question

(C) Given: Mass $m = 2 \ kg$, Potential energy $U = (12x + 16y) \ J$.The force acting on the ball is given by $\vec{F} = -
abla U = -\frac{\partial U

Vedclass · Accepted Answer

The speed of the ball at $t=2 \ s$ is $5 \ m/s$.

Vedclass · Answer

(C) Given: Mass $m = 2 \ kg$, Potential energy $U = (12x + 16y) \ J$.The force acting on the ball is given by $\vec{F} = -
abla U = -\frac{\partial U}{\partial x} \hat{i} - \frac{\partial U}{\partial y} \hat{j} = -12 \hat{i} - 16 \hat{j} \ N$.The acceleration of the ball is $\vec{a} = \frac{\vec{F}}{m} = \frac{-12 \hat{i} - 16 \hat{j}}{2} = (-6 \hat{i} - 8 \hat{j}) \ m/s^2$.Using the equation of motion $\vec{v}_f = \vec{v}_i + \vec{a}t$, where $\vec{v}_i = (15 \hat{i} + 20 \hat{j}) \ m/s$ and $t = 2 \ s$:$\vec{v}_f = (15 \hat{i} + 20 \hat{j}) + (-6 \hat{i} - 8 \hat{j}) 	imes 2$$\vec{v}_f = (15 \hat{i} + 20 \hat{j}) + (-12 \hat{i} - 16 \hat{j}) = (3 \hat{i} + 4 \hat{j}) \ m/s$.The speed at $t = 2 \ s$ is $|\vec{v}_f| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5 \ m/s$.Thus, the statement in option $C$ is correct.

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