A 25 kg block is initially at rest on a horizontal surface a...

A 2.5 kg block is initially at rest on a horizontal surface. A horizontal force F of magnitude 6.0 N and a vertical force P are then applied to the block. The coefficients for friction for the block and the surface are μs = 0.40 and μk = 0.25. Determine the magnitude of the frictional force acting on the block if the magnitude of P is (a) 8.0 N and (b) 12 N.

Transcript

00:01 Hello students in this question we are given that there is a block of mass 2 .5 kg is the block of mass 2 .5 kgs and on this block there are two forces acting the force one is acting in this direction horizontal force the value of this horizontal force is 6 newtons and there is a vertical force this is p okay now we have to calculate we are also given the coefficient of static friction which is equal to 0 .4 and the coefficient of kinetic friction which is 0 .25 okay and now we have two parts in this question part one and part two in both each part in part one the value of p is equals to the value of p in part one as given question it is 8 newton and in part two the value of p as given question it is equal to 12 newtons okay now we have to calculate what what will be the fractional force if p is 8 newtons okay so now let's solve that what will be the we have to find the magnitude of frictional force if the value of p is 8 newton's okay so first of all we have to find that whether the block will move or not for that we have to calculate the max value of maximum frictional force it will be equals to mu static multiplied by n where n is normal reaction okay so the value of static friction has given in question it is 0 .4 and the value of normal reaction will be equals to m g plus p so it will be equal to 0 .4 the value of mg will be 2 .5 multiplied by 10 plus the value of p as given good question it is 8 so it will be equal to 0 .4 multiplied after solving this we will get it will be 25 plus 8 which will be 33 and this will come out 13 .2 newtons so we can can see that this value is less than this value is less than six newtons which is horizontal force so the block will not move and the value of friction in this case will be equals to six newtons okay now let's solve the part two in part two we have to find again what will is the value of frictional force so for that we have to again find the value of f maximum so that we can see whether the block will move or not so it will also be equals to mu s multiplied by n value of mu s as a given question is to 0 .4 the n will be equals to m g plus p so it will be 0 .4 multiplied by 2 .5 multiplied by 10 plus the value of p in this question is 12 neutrons so it will be 0 .4 and this will be equals to 37 newtons okay so we can calculate 0 .4 multiplied by 37 is 14 .8 neutrons and this will be equal to 37...

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