If a #1 kg# object moving at #10 m/s# slows down to a halt after moving #50 m#, what is the coefficient of kinetic friction of the surface that the object was moving over?

To find the coefficient of kinetic friction, we can use the equation:

$f_k = \frac{m \cdot g \cdot d}{m \cdot g \cdot h}$

Where:

• $f_k$ is the coefficient of kinetic friction
• $m$ is the mass of the object (1 kg)
• $g$ is the acceleration due to gravity (approximately $9.81 \, \text{m/s}^2$)
• $d$ is the distance traveled by the object (50 m)
• $h$ is the height through which the object descended (not given)

However, since the object comes to a halt, we can infer that the work done by friction is equal to the initial kinetic energy of the object.

$f_k \cdot d = \frac{1}{2} m v^2$

Solve for $f_k$:

$f_k = \frac{\frac{1}{2} m v^2}{d}$

Plug in the given values:

$f_k = \frac{\frac{1}{2} \times 1 \, \text{kg} \times (10 \, \text{m/s})^2}{50 \, \text{m}}$

$f_k = \frac{50 \, \text{J}}{50 \, \text{m}}$

$f_k = 1 \, \text{N}$

Since the force of friction is equal to the weight of the object (1 kg), the coefficient of kinetic friction is 1.