Q. A block of mass `'m'` is sliding down a rough,fixed incline of andle `theta` with constant velocity.The coefficient of friction between the block and the incline is `mu.` The force that the block exerts on the incline is: A) zero B) `mg` C) `mgsintheta` D) `mu_k mgcostheta`
The forces acting on the block in the direction of incline are
`mgsintheta` and friction `muN` , where N is the normal force.
The forces acting on the block in the direction perpendicular to incline are
`mgcostheta` and N.
Since the acceleration is 0,
`mgsintheta-muN=0` and `mgcostheta-N=0` .
From here, `N=mgcostheta` and
`mgsintheta = muN = mumgcostheta`
So the friction coefficient must be `mu=sintheta/costheta=tantheta` .
Now consider the force that acting on the incline from the block. By the third Newton's Law, since there are two forces on the block from the incline - normal and friction - there are also two forces on the incline from the block, equal in magnitude and opposite in direction.
Thus, the force on the incline perpendicular to it equals `N=mgcostheta` .
The force on the incline parallel to it equal `muN = mumgcostheta` .
But since `mu = tantheta` , the parallel force is `tanthetamgcostheta=mgsintheta` .
The magnitude of NET force on incline from the block can be found as
`sqrt((mgcostheta)^2+(mgsintheta)^2) = mg` .
Choice B is correct.
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