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Consider the system shown in the figure. Block A has weight 4.91 N and block B has weight 2.94 N. Once block B is set into downward motion, it descends at a constant speed. Assume that the mass and friction of the pulley are negligible.

Calculate the coefficient of kinetic friction mu between block A and the table top.

I'm not sure where to begin.

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Kazeemsodikisola

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Explanation:

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The coefficient of kinetic friction between block A and the tabletop is obtained to be 0.6.

Two-body Problem

This problem can be analysed using the free body diagram of the two blocks. (Diagrams given as attachment.)

  • Consider the block 'B'. It descends at a constant speed. It has only vertical motion.
  • Also given, [tex]W_B=2.94\,N[/tex]

According to Newton's second law;

  • [tex]\sum F=ma[/tex]

Here, a = 0 m/s, therefore;

  • [tex]T-W_B=0[/tex]
  • [tex]\implies T = W_B = 2.94\,N[/tex]

  • Block A has a weight, [tex]W_A = 4.91N[/tex]

Now, consider the vertical forces in the case of block A.

The box has no vertical motion.

So, from Newton's second law;

  • [tex]\sum F=ma[/tex]
  • [tex]N-W_A=0[/tex]
  • [tex]\implies N=W_A=4.91\,N[/tex]

Now, consider the horizontal forces in the case of block A.

Here also the acceleration of the block is zero.

So, from Newton's second law;

  • [tex]\sum F=ma[/tex]
  • [tex]T-F_{fr}=0[/tex]
  • [tex]F_{fr}=T=2.94\,N[/tex]

But frictional force is given by;

  • [tex]F_{fr}= \mu _k N[/tex]
  • [tex]\implies \mu _k= \frac{F_{fr}}{N} =\frac{2.94}{4.91}=0.598[/tex]

[tex]\therefore \mu _k \approx 0.6[/tex]

Learn more about two-body problems here:

https://brainly.com/question/11341001

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