00:01
Well hello students now we're gonna see this question so basically this question has seven parts so first we'll start with part one okay an object has a coefficient of kinetic friction as 0 .2 so we need to find the kinetic friction force so fk which is the kinetic friction force is given by mu k multiplied by the normal force okay so mu k is what is given as 0 .2 multiplied by the normal okay so normal is given as 30 newton so this will come out to be what this will come out to be a 6 newton and this is only our answer of the first part.
00:32
Understood? so second part when we see, so second part is what? second part says that an object of coefficient of static friction is 0 .3 and normal force of 30 newton.
00:41
Okay, that means what we have to find coefficient of static friction or you can say force of static friction.
00:46
So it is mu s multiply by n.
00:48
Okay.
00:48
So this will be what? this will be mu s.
00:50
Mu as is what 0 .3 multiplied by n which is 30.
00:54
This will come out to be 9 newton.
00:56
Okay.
00:56
And this is the answer of the second part.
00:58
Now we'll see the third part.
01:00
Third part says that the object has 45 newton of static friction and the normal is 450 what is the coefficient of friction so basically we have fs and this is mu s multiply by n okay so what we have to do over here is that we have to we have been given fs okay so what is fs is basically 45 newton and mu as we have to find okay multiply by 450 which is the normal force okay so over here what we'll get we'll get mu s will be equal to do 45 by 450 so this will come out to be as what this will come out to be as 0 .1 okay so this is the muus okay so this is the answer of the third part now we'll call the fourth part when we see the fourth part what is the fourth part asking us there is a 80 newton of kinetic friction the coefficient is 0 .25 then what is the normal force so fk is given which is the force of kinetic friction mu k is given and and we have to find so from here what what will get normal force will be equal to fk divided by mu k okay so over here f is fk is basically 80 newton divided by the kinetic friction coefficient which is 0 .25 so this will come out to be as what this will come out to be as 320 newton understood now we'll proceed on to the fifth part so this is basically 320 newton is the answer of the fourth part now the fifth part has again three sections so we'll solve for part eight so an object has a mass of 20 kilogram and coefficient of friction as 0 .4 we have to find its fate so weight is what weight is basically equal to m multiply by g so m is what m is mass so mass is 20 and g we are going to take it as 9 .8 only so multiplied by 9 .8 so this will come out to be 196 newtons okay so this is the answer of the fifth part first part okay so we'll see about the second part of the fifth question so if the gravity and normal cancel then what is the normal force okay so basically over here when gravity and normal that means what mg must be equal to n okay so this will come out to be as what this will again come out to be 20 times okay so this is basically 20 multiplied by 9 .8 okay so we are going to see 20 multiplied by again 9 .8 because mg is what mg is only the gravitational force so this will again come out of be 196 mrs okay this is the answer of the second part of the fifth question now we'll see about the third part of the fifth question so the force of friction in this situation is what so force of friction or fs since the block is not moving okay so over you have still we have to consider that the block to be moving okay and what we have to say we have to say that the maximum friction possible will be what mu okay mu so this will write it as small f okay mu into n so mu is what? multiplyed by what is the normal force? normal force is 196.
04:07
Okay, so this will come out to be as 78 .4 newton's understood.
04:12
So this is the answer of the third part of the fifth question.
04:15
Now we'll see about the sixth question.
04:16
So the sixth question, sixth question says that what we have is that an object is known to have coefficient of friction mu k, mu k as 0 .17.
04:30
Okay so basically mu k over here is 0 .167 not one seven okay 167 so and the coefficient of static friction is also given mu s is given as 0 .42 okay and what we have to find if the normal force is 200 newton so we have to find how much frictional force will encounter while moving so while moving that means what only kinetic friction will act that means what fk will be equal to what since the object is moving so it will be equal to mu m k multiplied by m okay so mu k is what 0 .167 multiplied by 200 okay so this will come out to be at 33 .4 newton okay so this is the answer of the sixth question now we'll see about this last question which is the seventh question okay so the seventh question says that and 80 kilogram object of mu s is given so mu s is what mu s is basically 0 .35 okay and what we have mu k okay so basically over here here this is not mu s basically mu k is 0 .35 and mu s is also given it is 0 .6 okay and is assuming it on the flat surface what is the normal force of the object so basically over here since the mass is given as 80 kilogram okay so what will be in the normal force since the block is just resting on the on a plane okay that means for m g must be equal to n which is the normal force normal force will be equal to this is the first part okay m g okay so this m is what m is 80 multiplied by 9 .8 okay so this will come out as 784 okay so this will come out to be as 784 newton okay so this is the answer of the first part of the 7th question now we'll check about the second part of the 7th question how much force is required to get the object to start to move so to get the object to start to move fs max or you can say friction maximum amount of static friction must be overcome okay so what is the maximum amount of static friction it is mu s into n multiplied by n which is the normal force which we found in the first part okay so multiplied by 784 okay so this will come out to be as what this will come out to be as 470 .4 newton got it so this is the answer of the second part of the seventh question okay now we'll see about the last part of the last question if the above object is already moving and tension force 15 newton to the right is pulling it then what will be the net force on the object since since this object is already moving that means what over here fk will be applied that is the kinetic friction so what is kinetic friction it is mu k multiplied by n okay so this will be equal to 0 .35 multiplied by 784 newton so so this will come out to be 274 .4 newton...