A box of mass m is pulled up an incline that makes an angle...

A box of mass m is pulled up an incline that makes an angle ? with constant speed. If the box has traveled 2 meters, how much work is done; a) by the force F, b) by the force of friction, c) by gravity? d) Calculate the net work done on the box. Hint : Start by drawing a free body diagram and identify the forces acting on the box. m = 2 kg , ?k = 0.5 , g = 9.8 m/s^2 3 m 4 m

Transcript

00:01 Here in this given problem this is the inclined plane, height of the top of the inclined plane that is given as 3 meter and its base horizontal base that is 4 meter.

00:20 So, its hypotenuse means length of this plane using pythagoras theorem it will be calculated to be equal to 5 meter.

00:32 This is the box which is being pulled up the inclined, its mass is m.

00:44 So, its weight mg will be acting vertically down this is the angle theta made by the inclined plane with the horizontal component of this weight because this angle is also theta component of the weight perpendicular to the inclined plane that is mg cos theta along the inclined plane this is mg sin theta and the force being applied on the block to pull it up that is f.

01:17 This mass is given as 2 kilogram coefficient of kinetic friction 0 .5.

01:26 In the first part of the problem as the object is being pulled up.

01:38 So, the force of friction will be acting down along the inclined plane and this is the normal reaction n.

01:57 So, fk means force of friction will act down the inclined.

02:11 And as the object is moving up at constant speed means no acceleration hence no net force as the object is moving at constant speed.

02:37 So, the force acting up along the inclined that will be equal to forces some of the forces acting down along the inclined which include mg sin theta and force of friction fk.

02:57 And we know an expression for the force of friction that is mu k times the normal reaction and normal reaction we know using newton's third law of motion for every action there is an equal and opposite reaction.

03:12 So, that is mg sin theta plus mu k times n which is mg cos theta taking this mg as a common out leaving behind sin theta plus mu k cos theta finally, plugging in the known values 2 times of 9 .8 sin theta means perpendicular divided by hypotenuse plus mu k 0 .5 multiplied by base by hypotenuse.

03:54 And this force is calculated to be equal to 19 .6 newton and distance travelled over the inclined that is given to be 2 meter.

04:05 So, work done by this applied force that is fs into cos 0 degree because the force and distance are in the same direction.

04:16 So, angle between them that is 0 degree.

04:18 So, this is 19 .6 multiplied by 2 for cos 0 degree that is 1...

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