The potential energy in joules of a particle | vedclass

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Question

(D) The force acting on the particle is given by $\vec{F} = -
abla U = -(\frac{\partial U}{\partial x}\hat{i} + \frac{\partial U}{\partial y}\hat{j})

Vedclass · Accepted Answer

All of the above

Vedclass · Answer

(D) The force acting on the particle is given by $\vec{F} = -
abla U = -(\frac{\partial U}{\partial x}\hat{i} + \frac{\partial U}{\partial y}\hat{j}) = -3\hat{i} - 4\hat{j}\, N$.Given mass $m = 1\, kg$,the acceleration is $\vec{a} = \frac{\vec{F}}{m} = -3\hat{i} - 4\hat{j}\, m/s^2$.The magnitude of acceleration is $|\vec{a}| = \sqrt{(-3)^2 + (-4)^2} = 5\, m/s^2$. Thus,option $A$ is correct.To find when it crosses the $y$-axis $(x=0)$,we use $x(t) = x_0 + u_x t + \frac{1}{2} a_x t^2$. With $x_0 = 6$,$u_x = 0$,and $a_x = -3$,we have $0 = 6 - 1.5 t^2$,so $t^2 = 4$,which gives $t = 2\, s$.At $t = 2\, s$,the $y$-coordinate is $y(t) = y_0 + u_y t + \frac{1}{2} a_y t^2 = 4 + 0 - \frac{1}{2}(4)(2)^2 = 4 - 8 = -4$. Thus,option $C$ is correct.Using the work-energy theorem,$\Delta K + \Delta U = 0$. Initial energy $E_i = K_i + U_i = 0 + (3(6) + 4(4)) = 18 + 16 = 34\, J$. Final energy at $(0, -4)$ is $E_f = K_f + U_f = K_f + (3(0) + 4(-4)) = K_f - 16$. Since $E_i = E_f$,$34 = K_f - 16$,so $K_f = 50\, J$. Since $K = \frac{1}{2}mv^2$,$\frac{1}{2}(1)v^2 = 50$,so $v = 10\, m/s$. Thus,option $B$ is correct.Therefore,all statements are correct.

The potential energy in joules of a particle of mass $1\, kg$ moving in a plane is given by $U = 3x + 4y$,where the position coordinates $x$ and $y$ are measured in metres. If the particle is initially at rest at $(6, 4)$,then:

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