Q. 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.`

If a horizontal force `F` is applied on the block pressing it against the incline, it is seen that the block decelerates with retardation `a` . Find `a` .

A) `(Fcostheta)/m`

B) `(Fcostheta+muFsintheta)/m`

C) `(Fcostheta-muFsin^2theta)/m`

D) `(Fcos^2theta+muFsin^2theta)/m`

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**The correct choice is A.**

It is given that normally the block slides down the incline with the constant velocity, so the acceleration due to the friction and the component of gravity parallel to the incline is 0:

`F_f - mgsintheta=0` . ( `F_f` is the force of friction.)

Now the force F is applied horizontally, pressing the block against the incline, so it's component parallel to the incline is in the same direction as the friction: `Fcostheta` .

Now the second Newton's Law looks as follows:

`Fcostheta+F_f - mgsintheta = ma` . Here a is directed up the incline, so the block is decelerating.

Since `F_f - mgsintheta = 0` as stated above,

`Fcostheta = ma`

`a = (Fcostheta)/m` .

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