Question

6. The coefficient of friction of the slip between a body and the horizontal plane It is = 0,1. The body is released to slide with an initial velocity Vo = 2.6 m / s and after some time stops. a) Calculate the friction force of the slide when it is known that the body mass is 500 g. b) How long did the movement last until the body stopped?

Answer 1

a) We can find the magnitude of the friction force from the formula:

$$F_{k.fr.}=\mu_k mg=0.1\cdot 0.5\ kg\cdot 9.8\ \dfrac{m}{s^2}=0.49\ N.$$

b) Let's first find the deceleration of the body from the Newton's Second Law of Motion:

$$\sum F=ma,$$ $$F_{k.fr}=ma,$$ $$-\mu_k mg=ma,$$ $$a=-\mu_k g=-0.1\cdot 9.8\ \dfrac{m}{s^2}=-0.98\ \dfrac{m}{s^2}.$$

The sign minus means that the body decelerates.

Then, we can find the time that the body takes to finally stops:

$$v=v_0+at,$$ $$t=\dfrac{v-v_0}{a}=\dfrac{0\ \dfrac{m}{s}-2.6\ \dfrac{m}{s}}{-0.98\ \dfrac{m}{s^2}}=2.65\ s.$$

Answer:

a) $F_{k.fr.}=0.49\ N.$

b) $t=2.65\ s.$