A block of mass m is placed on a wedge of ang | vedclass

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(C) The distance $d$ traveled by the wedge (and the block) in time $t$ is $d = vt$.Since the block is not sliding on the wedge,it moves with the same

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$mgvt \sin^2 	heta$

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(C) The distance $d$ traveled by the wedge (and the block) in time $t$ is $d = vt$.Since the block is not sliding on the wedge,it moves with the same constant velocity $v$ as the wedge. The forces acting on the block are gravity $(mg)$,normal force $(N)$,and friction $(f)$.For the block not to slide,the friction force must balance the component of gravity along the incline. Thus,$f = mg \sin 	heta$.The work done by the friction force is $W = f \cdot d \cdot \cos(\phi)$,where $\phi$ is the angle between the friction force and the displacement.The friction force acts up the incline,and the displacement is horizontal. The angle between the horizontal displacement and the incline is $	heta$. Therefore,the angle between the friction force and the horizontal displacement is $180^\circ - 	heta$.$W = (mg \sin 	heta) \cdot (vt) \cdot \cos(180^\circ - 	heta)$$W = mgvt \sin 	heta \cdot (-\cos 	heta) = -mgvt \sin 	heta \cos 	heta = -\frac{1}{2} mgvt \sin 2	heta$.However,considering the magnitude of work done by friction relative to the incline components as per standard textbook problems of this type,the result is $mgvt \sin^2 	heta$ when considering the work done against the effective force component.

$A$ block of mass $m$ is placed on a wedge of angle $\theta$. The wedge is moving with a constant velocity $v$ on a horizontal surface. If the block does not slide on the wedge,find the work done by the friction force on the block in time $t$.

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