Calculating the Coefficient of Friction Between a Block and a Table

Problem Statement

To determine the coefficient of friction $\mu$ between a block and a horizontal table, consider the following setup: Two identical blocks of mass $m$ are used. One block is placed on a horizontal table, while the other is suspended off the edge, connected by a string that passes over a massless, frictionless pulley. The system starts from rest with both blocks at the same height $h$. When the hanging block falls to the floor, the block on the table slides an additional distance of $x = 2h$ before coming to a complete stop. Based on this information, calculate $\mu$.

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Approach 1: Using Kinematics

Hanging Block

The equation of motion for the hanging block is:

\[\sum F_y = m a_y \Rightarrow m g - F_T = m a_y\]

Where:

• $F_T$ is the tension in the string,
• $a_y$ is the acceleration of the block.

Table Block

The equation of motion for the block on the table is:

\[\sum F_x = m a_x \Rightarrow F_T - \mu m g = m a_x\]

Solve the System ($a = a_x = a_y$)

We solve for $a$ using the two equations:

1. $m g - F_T = m a$
2. $F_T - \mu m g = m a$

Adding these equations gives:

\[m g - \mu m g = 2 m a \Rightarrow a = \frac{1}{2} g (1 - \mu)\]

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Final Velocity After Falling Height $h$

Using the kinematic equation for the falling block:

\[v_h^2 = v_i^2 + 2 a \Delta x \Rightarrow v_h^2 = 0 + 2 \left(\frac{1}{2} g (1 - \mu)\right) h\] \[v_h = \sqrt{g h (1 - \mu)}\]

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Deceleration on the Table After the Hanging Block Hits the Ground

After the hanging block hits the floor, the block on the table decelerates due to friction. The equation of motion is:

\[\sum F_x = m a_x \Rightarrow -\mu m g = m a_x \Rightarrow a_x = -\mu g\]

Using the kinematic equation:

\[0 = v_h^2 + 2 a \Delta x \Rightarrow 0 = g h (1 - \mu) + 2 (-\mu g) (2h)\]

Simplifying:

\[0 = g h (1 - \mu) - 4 \mu g h \Rightarrow 0 = g h (1 - 5 \mu)\]

Solving for $\mu$:

\[\mu = \frac{1}{5}\]

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Approach 2: Using Kinetics

First Stage (Falling Height $h$)

The work-energy principle relates the work done by friction to the change in energy:

\[W_f = \Delta E \Rightarrow -f h = (K_f + U_f) - (K_i + U_i)\]

Substituting the forces and energies:

\[-k \mu m g h = \left(\frac{1}{2} m v_h^2 + \frac{1}{2} m v_h^2 + m g h\right) - \left(0 + m g h + m g h\right)\]

Simplifying:

\[-k \mu m g h = m v_h^2 - m g h\]

Using $v_h^2 = g h (1 - \mu)$:

\[-k \mu m g h = m \left(g h (1 - \mu)\right) - m g h\]

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Second Stage (Sliding Distance $2h$)

For the block sliding on the table, the work-energy principle is:

\[W_f = \Delta E \Rightarrow -f (2h) = (K_f + U_f) - (K_i + U_i)\]

Substituting the forces and energies:

\[-2 \mu m g h = \left(0 + m g h\right) - \left(\frac{1}{2} m v_h^2 + m g h\right)\]

Using $v_h^2 = g h (1 - \mu)$:

\[-2 \mu m g h = -\frac{1}{2} m g h (1 - \mu)\]

Simplifying:

\[\mu = \frac{1}{5}\]

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Conclusion

Using both kinematics and kinetics, we find that the coefficient of friction is:

\[\mu = \frac{1}{5}\]

This consistent result demonstrates how energy principles and force analysis can complement each other in solving physics problems.