Answer to Question #102182 in Mechanics | Relativity for dfd

Question

A block with an initial velocity of 8.9 m/s slides across a rough surface until it eventually comes to rest. If the coefficient of friction is equal to 0.83, then how far will the block travel?

Answer 1

We can find the distance that the block travel until it eventually comes to rest from the kinematic equation:

$v_f^2 - v_i^2 = 2as,$

here, $v_i = 8.9 \dfrac{m}{s}$ is the initial velocity of the block, $v_f = 0 \dfrac{m}{s}$ is the final velocity of the block, $a$ is the acceleration of the block and $s$ is the distance that the block travel until it eventually comes to rest.

From this equation we can find the distance that the block travel until it eventually comes to rest:

$s = \dfrac{-v_i^2}{2a}.$

We can find the acceleration of the block from the Newton's Second Law of Motion:

$\sum F_x = ma_x$ $-F_{fr} = ma,$ $a = \dfrac{-F_{fr}}{m} = \dfrac{- \mu N}{m} = \dfrac{- \mu mg}{m} = - \mu g.$

Finally, substituting expression for $a$ into the formula for distance, we get:

$s = \dfrac{-v_i^2}{2(- \mu g)} = \dfrac{(8.9 \dfrac{m}{s})^2}{2 \cdot 0.83 \cdot 9.8 \dfrac{m}{s^2}} = 4.87m.$

Answer:

$s = 4.87m.$