Can Someone Help Of With Question Work Energy And Power

How do I solve this mechanics question on work energy power?

Refer to the diagram below. By the law of conservation of energy, [math]\frac{1}{2}mV^{2}=mgl(1-\cos\theta)+\frac{1}{2}mv^{2}[/math]. We also have [math]\frac{mv^{2}}{l}=T-mg\cos\theta[/math], where [math]T[/math] is the tension. Let the angle at which the string becomes slack i.e., [math]T=0[/math], be [math]\alpha[/math]. Then, the first and second equations give [math]\cos\alpha=\frac{2}{3}(1-\frac{V^{2}}{2gl})[/math] and the speed of the particle at the angle [math]\alpha[/math] is given by [math]v^{2}=-\frac{2}{3}gl(1-\frac{V^{2}}{2gl})[/math]. The unit vectors along the string and in the direction of increasing [math]\theta[/math] are: [math]\sin\theta\hat{i}+\cos\theta\hat{j}[/math] and [math]\cos\theta\hat{i}-\sin\theta\hat{j}[/math] respectively. So the position of the particle at any angle [math]\theta[/math] is [math](l\sin\theta, l\cos\theta)[/math] and the velocity is [math](v\cos\theta, -v\sin\theta)[/math].At the angle [math]\alpha[/math] the particle begins to proceed on a parabola, so that the only force acting on it is its weight. The equations of motion then are : [math]m\frac{d^{2}x}{dt^{2}}=0[/math], [math]m\frac{d^{2}y}{dt^{2}}=mg[/math], or [math]\frac{d^{2}x}{dt^{2}}=0[/math] and [math]\frac{d^{2}y}{dt^{2}}=g[/math] (Note the direction of y in the diagram). We solve these equations using the initial conditions for the position and velocity of the particle which are [math]x=l\sin\alpha[/math], [math]y=l\cos\alpha[/math], [math]v_{x}=v\cos\alpha[/math], [math]v_{y}=-v\sin\alpha[/math], where [math]v^{2}=-\frac{2}{3}gl(1-\frac{V^{2}}{2gl})[/math], as seen above. The solutions are [math]x=(v\cos\alpha)t+l\sin\alpha[/math], [math]y=\frac{1}{2}gt^{2}-(v\sin\alpha)t+l\cos\alpha[/math]. Since the particle passes through the point of suspension [math]O(0,0)[/math], let at time t, the position [math](x,y)[/math] of the particle be [math](0,0)[/math]. Then, from the equation for x we get [math]t=-\frac{l\sin\alpha}{v\cos\alpha}[/math] which upon substitution in the equation for y gives [math]\frac{1}{2}\left(-\frac{l\sin\alpha}{v\cos\alpha}\right)^{2}-(v\sin\alpha)\left(-\frac{l\sin\alpha}{v\cos\alpha}\right)+l\cos\alpha=0[/math]. After a certain amount of algebra, this equation gives [math]V^{2}=2gl(1\pm\frac{\sqrt{3}}{2})[/math]. The minus sign is ignored as this value of V causes the particle to oscillate. The value of V that leads to parabolic motion is given by [math]V^{2}=2gl(1+\frac{\sqrt{3}}{2})[/math].

How do I solve this question based on forces, work and energy in physics?

Relative velocity of the rod and inclined plane along the direction perpedicular to the surface of incline plane will be 0 to maintain contact. You can find the constraint between the velocities of both bodies using this. Then just conserve the mechanical energy of the system initially and finally as there is no non conservative force acting on the system. Work done by the normal force on the rod by the clamps is also zero as it is perpendicular to the direction of motion.Answer is-​​