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Q. A block of mass `'m'` is sliding down a rough, fixed incline of angle `theta` with...

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user8235304 | Student, Grade 11 | Valedictorian

Posted September 10, 2013 at 5:57 PM via web

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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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ishpiro | Teacher | (Level 1) Associate Educator

Posted September 10, 2013 at 7:31 PM (Answer #2)

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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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