# 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`

Images:
This image has been Flagged as inappropriate Click to unflag 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,...

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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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