Question

The wooden box is thrown with some initial speed in the horizontal plane and after crossing the twilight 5 m for 2 s it stops. Calculate the coefficient of sliding friction.

Answer 1

Let's first find the initial velocity of the box:

$$d=\dfrac{v_0+v}{2}t,$$ $$v_0=\dfrac{2d}{t}-v=\dfrac{2\cdot 5\ m}{2\ s}-0=5\ \dfrac{m}{s}.$$

Then, we can find the deceleration of the box:

$$v=v_0 + at,$$ $$a=\dfrac{v-v_0}{t}=\dfrac{0\ \dfrac{m}{s}-5\ \dfrac{m}{s}}{2\ s}=-2.5\ \dfrac{m}{s^2}.$$

The sign minus means that the box decelerates.

Finally, from the Newton's Second Law of Motion we can find the coefficient of sliding friction:

$$\sum F=ma,$$ $$F_{k.fr.}=ma,$$ $$-\mu_k mg=ma,$$ $$\mu_k=-\dfrac{a}{g}=-\dfrac{-2.5\ \dfrac{m}{s^2}}{9.8\ \dfrac{m}{s^2}}=0.25$$

Answer:

$\mu_k=0.25$