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A box of mass 20 kg is at rest on a rough horizontal floor. The coefficient of friction...

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saj-94 | Salutatorian

Posted August 3, 2013 at 12:54 PM via web

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A box of mass 20 kg is at rest on a rough horizontal floor. The coefficient of friction between the box and the floor is 0·3. The box is subjected to a horizontal force of TN.

Given that T = 65, find the frictional force and magnitude of the acceleration of the box

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llltkl | College Teacher | Valedictorian

Posted August 3, 2013 at 3:19 PM (Answer #1)

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Refer to the attached image for the given situation.

Here,

`SigmaF_y=0`

`rArr F_(norm)-F_(grav)=0`

`rArr F_(norm)=F_(grav)=m*g=20*9.81=196.2 N`

Again, along the x-direction,

F_(net)=m*a=F_(app)-F_(fric)

F_(fric)=coefficient of friction*F_(norm)

`=0.3*196.2=58.86 N`

Acceleration, a = F_(net)/m

`=(65-58.86)/20`

`=0.307` m/s^2

Therefore, the frictional force on the surface is 58.86 N and magnitude of the acceleration of the box is 0.307 m/s^2.

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