Question

A spring with spring constant of 175 N/m is used to pull a 10 kg crate across a level floor. If the coefficient of friction between the crate and the ground is .3 and the crate gets an acceleration of 4.5 m/s2, how far does the spring get stretched?

Answer 1

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

$$\sum F_x=ma_x,$$ $$F-F_{fr}=ma,$$ $$kx-\mu mg=ma.$$

From this equation we can find the elongation of the string:

$$x=\dfrac{m(\mu g+a)}{k},$$ $$x=\dfrac{10\ kg\cdot(0.3\cdot 9.8\ \dfrac{m}{s^2}+4.5\ \dfrac{m}{s^2})}{175\ \dfrac{N}{m}}=0.42\ m.$$

Answer:

$x=0.42\ m.$