Question

A 7.5 kg object on a rough surface with coefficient of kinetic friction 0.1 is pulled by a constant tension 60 N directly to the right. If the mass started at rest, how far does it go after 5 s

 seconds?

Answer 1

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

$$F_T-F_{fr}=ma,$$ $$F_T-\mu_k mg=ma,$$ $$a=\dfrac{F_T-\mu_k mg}{m},$$ $$a=\dfrac{60\ N-0.1\cdot7.5\ kg\cdot9.8\ \dfrac{m}{s^2}}{7.5\ kg}=7.02\ \dfrac{m}{s^2}.$$

Then we can find the velocity of the object after 5 seconds:

$$v=v_0+at=0+7.02\ \dfrac{m}{s^2}\cdot5\ s=35.1\ \dfrac{m}{s}.$$

Finally, we can find the distance the object traveled after 5 seconds from the kinematic equation:

$$v^2=v_0^2+2ad,$$ $$d=\dfrac{v^2}{2a}=\dfrac{(35.1\ \dfrac{m}{s})^2}{2\cdot7.02\ \dfrac{m}{s^2}}=87.75\ m.$$