Questions: Dexter Eius is running through the cafeteria when he slips on some mashed potatoes and falls to the floor. Dexter lands in a puddle of milk and skids to a stop with an acceleration of 5.6 m / s / s. Dexter weighs 703 Newtons. Determine the coefficient of friction between Dexter and the milky floor.

Solution Steps

We are given the following values:

• Acceleration, \( a = 5.6 \, \mathrm{m/s^2} \)
• Weight of Dexter, \( W = 703 \, \mathrm{N} \)

The weight \( W \) is related to the mass \( m \) by the equation: \[ W = m \cdot g \] where \( g \) is the acceleration due to gravity (\( g \approx 9.81 \, \mathrm{m/s^2} \)).

Rearranging to solve for \( m \): \[ m = \frac{W}{g} = \frac{703 \, \mathrm{N}}{9.81 \, \mathrm{m/s^2}} \approx 71.659 \, \mathrm{kg} \]

The frictional force \( F_f \) is what causes Dexter to decelerate. According to Newton's second law: \[ F_f = m \cdot a \] Substituting the values: \[ F_f = 71.659 \, \mathrm{kg} \times 5.6 \, \mathrm{m/s^2} \approx 401.2904 \, \mathrm{N} \]

The normal force \( F_N \) is equal to Dexter's weight because he is on a horizontal surface: \[ F_N = W = 703 \, \mathrm{N} \]

The coefficient of friction \( \mu \) is given by the ratio of the frictional force to the normal force: \[ \mu = \frac{F_f}{F_N} = \frac{401.2904 \, \mathrm{N}}{703 \, \mathrm{N}} \approx 0.5707 \]

Final Answer