A block with an initial speed of 10 m/s slides on a horizontal surface and comes to rest after traveling a distance of 25 m. What is the coefficient of kine...

Answer Details

When a block slides on a horizontal surface and comes to rest, the only horizontal force acting on it is the kinetic friction force. This friction force produces a deceleration that brings the block to a stop.

The friction force on a horizontal surface is given by:

\[ f = \mu_k m g \]

where \( \mu_k \) is the coefficient of kinetic friction, \( m \) is the mass of the block, and \( g \) is the acceleration due to gravity. By Newton's second law, the deceleration \( a \) equals \( \mu_k g \).

Using the kinematic equation for motion with constant deceleration:

\[ v^2 = u^2 - 2as \]

The block starts at \( u = 10 \) m/s and comes to rest (\( v = 0 \)) after travelling \( s = 25 \) m. Substituting:

\[ 0 = (10)^2 - 2 \times a \times 25 \]

\[ 0 = 100 - 50a \]

\[ a = \frac{100}{50} = 2 \text{ m/s}^2 \]

Since \( a = \mu_k g \):

\[ \mu_k = \frac{a}{g} = \frac{2}{9.8} \approx 0.204 \]

Rounding to one decimal place, the coefficient of kinetic friction is approximately 0.2.

A common mistake is to forget that the deceleration on a horizontal surface due to friction depends only on \( \mu_k \) and \( g \), not on the mass of the block (mass cancels out). This is why the question does not need to provide the mass.