Answer to Question #164636 in Mechanics | Relativity for Lucy green

Question

A 15.0 kg child slides down a 4.25 m playground slide that makes a 45o with the horizontal axis. While sliding, a 2.5 N force of friction acts on the child. What will the child’s speed be when she reaches the bottom of the slide?

Answer 1

Let's apply the Newton's Second Law of Motion and find the acceleration of the child:

$$mgsin\theta-F_{fr}=ma,$$ $$a=\dfrac{mgsin\theta-F_{fr}}{m},$$ $$a=\dfrac{15\ kg\cdot9.8\ \dfrac{m}{s^2}\cdot sin45^{\circ}-2.5\ N}{15\ kg}=6.76\ \dfrac{m}{s^2}.$$

Finally, we can find the child’s speed when she reaches the bottom of the slide from the kinematic equation:

$$v^2=v_0^2+2ad,$$ $$v=\sqrt{v_0^2+2ad},$$ $$v=\sqrt{0+2\cdot6.76\ \dfrac{m}{s^2}\cdot4.25\ m}=7.58\ \dfrac{m}{s}.$$