A smooth right-angled wedge of mass M is kept | vedclass

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(B) Let $V_w$ be the velocity of the wedge of mass $M$ when the block of mass $m$ reaches the ground. Since there is no external horizontal force on t

Vedclass · Accepted Answer

$\frac{-1}{2} Mv^2$

Vedclass · Answer

(B) Let $V_w$ be the velocity of the wedge of mass $M$ when the block of mass $m$ reaches the ground. Since there is no external horizontal force on the system, the horizontal component of the momentum of the system is conserved.Initially, the system is at rest, so the total initial horizontal momentum is $0$.Thus, $m v_x + M V_w = 0$, where $v_x$ is the horizontal component of the velocity of mass $m$. This implies $v_x = -\frac{M V_w}{m}$.By the work-energy theorem for the block $m$, the work done by gravity plus the work done by the normal force $N$ equals the change in kinetic energy of $m$.$W_g + W_N = \frac{1}{2} m V^2$.Since $W_g = mgh$, we have $W_N = \frac{1}{2} m V^2 - mgh$.Alternatively, consider the wedge. The only horizontal force on the wedge is the normal force from the block. The work done by the normal force on the wedge is $W_{N'} = \Delta K_w = \frac{1}{2} M V_w^2$.Since the normal force on the block is equal and opposite to the normal force on the wedge, and the displacement of the block is relative to the wedge, we use the work-energy theorem on the wedge: $W_{N'} = \int N \cdot dx_w = \frac{1}{2} M V_w^2$.The work done by the normal force on the block is $W_N = -\int N \cdot dx_{rel}$.Using the conservation of energy for the system: $mgh = \frac{1}{2} m V^2 + \frac{1}{2} M V_w^2$.From $m v_x + M V_w = 0$, we have $V_w = \frac{m |v_x|}{M}$.For a smooth wedge, the work done by the normal force on the block is equal to the negative of the kinetic energy gained by the wedge: $W_N = -\frac{1}{2} M V_w^2$.Substituting $V_w = \frac{m v_x}{M}$ is complex, but using the relation $W_N = -\Delta K_{wedge}$, we find the correct expression is $W_N = -\frac{1}{2} M V_w^2$. Given the options, the correct answer is $B$.

$A$ smooth right-angled wedge of mass $M$ is kept on a smooth horizontal surface as shown. $A$ mass $m$ is released from the top of the wedge. When $m$ reaches the ground, its speed is $V$. Find the work done by the normal contact force on $m$ while it comes down to the ground.

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