A 620 kg block sits on a frictionless tabletop a 230 kg cube...

A 62.0 kg block sits on a frictionless tabletop. A 23.0 kg cube sits on top of the block. The cube is being pulled to the right by a horizontal force with a 150 N magnitude (Figure 6-39). The coefficient of static friction between the block and the cube is 0.400 and the coefficient of kinetic friction is 0.250. Assume all the quantities have 3 significant digits. (a) Is the static friction between the two blocks strong enough to cause them to stick together when the horizontal force acts? blank1 - Word Answer (yes/no) (b) What is the acceleration of the cube? blank2 - Numeric Answer m/s^2 to the right. (c) What is the acceleration of the block? blank3 - Numeric Answer m/s^2 to the right.

Transcript

00:01 In this question we have given a 62 kg block which is on a frictionless table and it is told that a 23 kg cube is on the top of this block.

00:11 Now it is told that the cube is being pulled by a force of 150 newton and we have given the coefficient of static friction between block and cube and cofficient of kinetic friction is given to us.

00:24 Now in the first part of the problem we have to tell that is the static friction between the two blocks is strong in a to cause the stick together when the horizontal force is acting on this so first of all i am writing the given data for this question so we have given that mass of block that is capital m is equal to this is 62 kg and we have given the mass of cube that is small m and this is 23 kg we have given the coefficient of stative friction that is mu as is equal to this is 0 .4 and kinetic friction is that is 0 .250.

01:16 Now if we draw the diagram for this question so this is our block of mass 62 kg and there is a cube on this block so this mass is small and this is capital m and we had apply a horizontal force on this cube that is f is equal to 150 newton so we can say that normal force or normal contact force on the qbs that is n1 is this is m into g so we can say if we find the value of static friction so we can say that is static friction force that is f s is equal to this is mu s into n1 so we can say this will comes out to be now mu s is 0 .4 into n1 is this is 23 into 9 .81 so this is equal to 90 0 .3 newton.

02:33 So we can say that since the applied force that is f is greater than the value of static friction force that is the limiting value so we can say that the cube will not be at rest on the block so kinetic friction will act.

03:13 So in the first part of the problem it is asked we have to tell that is the two blocks are stick together so we can say that no, they both are not stick together.

03:28 So we can say that for the first part of the problem our answer is no because f is greater than the static friction that is f s now in the second part of the problem we have to find the acceleration of the cube so we can say that if we draw the free body diagram of the cube so this is the cube here and mass of the cube is m and we had apply a force on this cube that is f here and and there is a friction acting on this cube that is kinetic friction that is fk and let the acceleration of this block in this direction is a1 so we can say that and the normal force acting on this cube is that is n1 and the weight of this cube is acting vertically downward that is m g so we can say that the kinetic friction force that is fk is equal to this is mu k into n1 so we can say from newton's second law that is net force is a equals to mass into acceleration that is m into a 1 so net force in horizontal direction is that is applied force f minus kinetic friction force f k and this is equals to m into a 1 so from here we get the value of acceleration so this is 150 minus this is 0 .25 into 23 into this is 9 .81 and mass is 23 into this is a 1 so from here we get the value of acceleration a 1 and this is equals to 4 .07 meter per second square.

05:13 So this is the acceleration of the cube...

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