Answer to Question #159701 in Mechanics | Relativity for Dr. Horus

Question

A 12 kg mass was projected with

an initial kinetic energy of 450J along

the surface of a rough plane inclined

at 30° above the horizontal. If the

mass comes to rest after traveling 3.5m along the plane, calculate the

coefficient of kinetic friction between the mass and the plane.

Answer 1

Let's apply the Newton's Second Law of Motion:

$$-mgsin\theta-\mu_kmgcos\theta=ma.$$

From this equation we can find the coefficient of kinetic friction between the mass and the plane:

$$\mu_k=\dfrac{-gsin\theta-a}{gcos\theta}.$$

Let's find the initial velocity of the mass from the definition of kinetic energy:

$$v=\sqrt{\dfrac{2KE}{m}}=\sqrt{\dfrac{2\cdot450\ J}{12\ kg}}=8.66\ \dfrac{m}{s}.$$

Then, we can find the deceleration of the mass:

$$a=\dfrac{v^2-v_0^2}{2d}=\dfrac{0-(8.66\ \dfrac{m}{s})^2}{2\cdot3.5\ m}=-10.71\ \dfrac{m}{s}.$$

Finally, substituting $a$ into the formula for $\mu_k$, we get:

$$\mu_k=\dfrac{-9.8\ \dfrac{m}{s^2}\cdot sin30^{\circ}-(-10.71\ \dfrac{m}{s})}{9.8\ \dfrac{m}{s^2}\cdot cos30^{\circ}}=0.68$$