Answer to Question #168404 in Physics for MJ

Question

A 2.00-kg toy accelerates from rest to 3.00 m/s in 8.00 s on rough surface of uk=0.300. Find the applied force f

Answer 1

Let's first find the acceleration of the toy:

$$a=\dfrac{v-v_0}{t}=\dfrac{3.0\ \dfrac{m}{s}-0}{8.0\ s}=0.375\ \dfrac{m}{s^2}.$$

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

$$F_{appl}-\mu_kmg=ma,$$ $$F_{appl}=m(a+\mu_kg),$$ $$F_{appl}=2.0\ kg\cdot(0.375\ \dfrac{m}{s^2}+0.3\cdot9.8\ \dfrac{m}{s^2})=6.63\ N.$$