00:01
Hello everyone in this question it is given that mark m1 equals to m2 equals to 2 gig and here aesthetic friction between mark m1 and the table is given by nu f equals to 0 .40 and proficient of kinetic question nuke equals to given 0 .30 so here we need to find out both part a and part b minimum value of so now let's see the answer so for part air for the block m2 if the mass m2 is not moving or in equilibrium conditions then we can retain as b equal to m2 multiplied by v so here we can take the value of g equals to 9 .8 meter apart second square so now we will put the values therefore we will got 2 multiplied by 9 .8 so after solving this we will got the value of t equals to 19 .6 newton.
01:07
So, the mass m1, we know that summation of x equals to 0, which is horizontal force equals to 0.
01:17
Therefore, we will got tension t equals to seasonal force, mu s multiplied by normal force.
01:24
Therefore, we will got mu s multiplied by m1 multiplied by z, multiplied by z.
01:33
N equals to m1 multiplied by g so now here we will put the value therefore we will find the value of m1 so now we can attain as m1 equals to t divided by new s multiplied by g so the value of t is 19 .6 divided by the value of new is the value of new is the value of new f is 0 .4 multiplied by 9 .8 so after solving this we will go the value of m1 equals to 5 quiv so minimum value of m1 will keep the system from starting to move is 5 pb so now coming to part b if the system is moving with constant state then explanation is 0 for the block m2 if the mask m2 is not moving or in equilibrium we are so, if the system is moving with constant speed, then acceleration is 0.
02:39
So for the block m2, if the mark m2 is not moving or in equilibrium conditions, so we can detain as t equals to m2 multiplied by z.
02:49
So from here we will got t equals to 19 .6 newton.
02:56
Again we know that for mass m1, summation of x equals 0, since it is in equilibrium...