Question

A 6.0 𝑙𝑏 box is pulled along the horizontal floor by a rope that makes an angle of 30o above the horizontal. The coefficient of kinetic friction between the box and the floor is 0.10. If the tension in the rope is 1.0 𝑙𝑏, find the acceleration of the box. 

Answer 1

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

$$Tcos\theta-\mu W=ma,$$ $$a=\dfrac{Tcos\theta-\mu W}{m},$$ $$a=\dfrac{1.0\ lb\cdot cos30^{\circ}-0.10\cdot6.0\ lb}{\dfrac{6.0\ lb}{32.17\ \dfrac{ft}{s^2}}}=1.426\ \dfrac{ft}{s^2},$$ $$a=1.426\ \dfrac{ft}{s^2}\cdot\dfrac{1\ m}{3.28084\ ft}=0.43\ \dfrac{m}{s^2}.$$

Answer:

$a=1.426\ \dfrac{ft}{s^2}=0.43\ \dfrac{m}{s^2}.$