A bartender slides a glass mug with a mass of 520 g along the top of a level bar. The coefficient of kinetic friction between the glass and the bar is 0.09. When the glass leaves the bartender's hand, it is moving, it is moving at a speed of 3.9 m/s. How far does the glass travel before coming to a stop?
Use SI units in your answer.
\Delta x =
To calculate the distance the glass travels before coming to a stop, we need to determine the deceleration force acting on the glass. We can then use this force to calculate the distance using the equation of motion.
The deceleration force can be found using the equation:
\[ F_{\text{decel}} = \mu_k \cdot m \cdot g \]
where
\( F_{\text{decel}} \) is the deceleration force,
\( \mu_k \) is the coefficient of kinetic friction,
\( m \) is the mass of the glass,
and \( g \) is the gravitational acceleration (approximately 9.8 m/s^2).
Substituting the given values into the equation:
\[ F_{\text{decel}} = 0.09 \cdot 0.52 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 \]
\[ F_{\text{decel}} = 0.04644 \, \text{N} \]
Next, we need to calculate the acceleration of the glass using the equation:
\[ a = \frac{F_{\text{decel}}}{m} \]
Substituting the values:
\[ a = \frac{0.04644 \, \text{N}}{0.52 \, \text{kg}} \]
\[ a = 0.089 \, \text{m/s}^2 \]
Now, we can use the equation of motion to calculate the distance traveled by the glass before coming to a stop:
\[ \Delta x = \frac{v_f^2 - v_i^2}{2a} \]
where
\( \Delta x \) is the distance traveled by the glass,
\( v_f \) is the final velocity (which is 0 m/s as the glass comes to a stop),
and \( v_i \) is the initial velocity (which is 3.9 m/s).
Substituting the values:
\[ \Delta x = \frac{0 - (3.9 \, \text{m/s})^2}{2 \cdot 0.089 \, \text{m/s}^2} \]
\[ \Delta x = \frac{-15.21}{0.178} \]
\[ \Delta x = -85.44 \, \text{m} \]
Since distance cannot be negative, we take the absolute value of the result:
\[ \Delta x = 85.44 \, \text{m} \]
Therefore, the glass will travel approximately 85.44 meters before coming to a stop.