# 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`

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Inna Shpiro | Certified Educator

calendarEducator since 2013

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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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## Related Questions

llltkl | Student

A block of mass m is sliding down a rough, fixed incline of angle `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 its own weight, mg which operates in the downward direction. This can be resolved into two mutually perpendicular components, though their vector summation will be the same, i.e. `mg` .

Therefore, option B) has the correct answer.