Two blocks, stacked one on top of the other, slide on a frictionless horizontal surface. The surface between the two blocks is rough, however, with a coefficient of static friction equal to 0.40. The top block has a mass of 2.9 kg, and the bottom block's mass is 5.8 kg. If a horizontal force F is applied to the bottom block, what is the maximum value F can have before the top block begins to slip?

The value is

Explanation:

From the question we are told that

    The coefficient of static friction is  

    The mass of the top of block  is  

     The mass of the  bottom is  

Generally given that the frictional force between the bottom block and the frictionless horizontal is zero then the maximum value F can have before the top block begins to slip will be equal to the frictional force between the top and  the bottom block which is mathematically represented as

Here N is the normal force acting on top block which is mathematically represented as

=>

So

=>  

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.