Question

A clerk moves a box of cans down the aisle by pulling on a strap attached to the box. The clerk pulls with a force of 185.0 N at an angle of 25.0° with the horizontal. The box has a mass of 35.0 kg and the coefficient of friction between the box and the floor is 0.450. Find the acceleration of the box.

Answer 1

Givens.

• The pulling force 185 N.

,

• The angle is 25 degrees.

,

• The mass is 35 kg.

,

• The coefficient of friction is 0.45.

Let's use Newton's Second Law to find an expression.

$\begin{gathered} \Sigma F_x=ma\to F\cdot\cos \theta-\mu mg=ma \\ 185N\cdot\cos 25-0.45\cdot35\operatorname{kg}\cdot9.81\cdot\frac{m}{s^2}=ma \\ 167.67N-154.51N=35\operatorname{kg}\cdot a \\ a=\frac{13.16}{35}\cdot\frac{m}{s^2} \\ a=0.38\cdot\frac{m}{s^2} \end{gathered}$

Therefore, the acceleration of the box is 0.38 m/s^2.