Question

a 3kg box slides down a 30.0 degree ramp with an acceleration of 1.4 m/sec2. Determine the coefficient of kinetic friction between the box and the ramp.

Answer 1

Question#5017

a 3kg box slides down a 30.0 degree ramp with an acceleration of $1.4 \, \text{m/sec}^2$ . Determine the coefficient of kinetic friction between the box and the ramp.

Solution:

Let:

$$m = 3 K g$$ $$\alpha = 30{}^{\circ}$$ $$a = 1.4 \, \text{m/sec}^2$$

$n = ?$ coefficient of kinetic friction

$$n = \frac{F(friction)}{F(L)}; F(friction) = n * F(L)$$ $$F(L) = mg * \sin \alpha$$ $$a = \frac{F(L) - F(friction)}{m} \text{ (Newton's second law)}$$ $$a = \frac{F(L) - n * F(L)}{m}$$ $$a = \frac{F(L) * (1 - n)}{m}$$ $$1 - n = \frac{a m}{F(L)}$$ $$n = 1 - \frac{a m}{m g \sin \alpha}; \, n = 1 - \frac{a}{g \sin \alpha}$$ $$n = 1 - \frac{1.4}{9.8 * 0.5} = 0.714$$

Answer: 0,714