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A large block is being pushed against a smaller block such that the smaller block remains elevated while being pushed. The mass of the smaller block is m = 0.45 kg. It is found through repeated experimentation that the blocks need to have a minimum acceleration of a = 13 m / s 2 in order for the smaller block to remain elevated and not slide down. What is the coefficient of static friction between the two blocks?

Respuesta :

Explanation:

According to the free body diagram a block of mass m will have expression for force as follows.

                  N = ma

and,       [tex]f_{c} - mg[/tex] = 0  

             [tex]\mu_{s}N - mg[/tex] = 0

       [tex]\mu_{s} = \frac{mg}{N}[/tex] = [tex]\frac{mg}{ma}[/tex]

                  = [tex]\frac{g}{a}[/tex]

                  = [tex]\frac{9.8}{13}[/tex]

                  = 0.75

Therefore, we can conclude that the value of coefficient of static friction between the two blocks is 0.75.

Ver imagen OlaMacgregor
Lanuel

The coefficient of static friction between the large block and smaller block is equal to 0.754.

Given the following data:

  • Mass of smaller block (m) = 0.45 kg.
  • Acceleration (a) = [tex]13 \;m/s^2[/tex]

Acceleration due to gravity = 9.8 [tex]m/s^2[/tex]

To determine the coefficient of static friction between the large block and smaller block:

A force of static friction can be defined as the frictional force that resists the relative motion of two (2) surfaces.

Hence, a force of static friction is a frictional force that keeps an object at rest or stationary rather than being in relative motion.

Mathematically, the force of static friction is given by the formula;

[tex]Fs = uFn[/tex]

Where;

  • Fs represents the force of static friction.
  • μ represents the coefficient of friction.
  • Fn represents the normal force.

For these block systems, the forces acting on them is given by:

[tex]uma - mg = 0\\\\uma = mg\\\\ua =g\\\\u=\frac{g}{a}[/tex]

Substituting the parameters into the formula, we have;

[tex]u=\frac{9.8}{13}[/tex]

u = 0.754

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