A \(10.0 kg\) mass is dragged along a horizontal table by a \(60 N\) force. The force is inclined at angle \(30^{\circ}\) to the horizontal, and the coefficient of dynamic friction between the mass and the table is \(0.15\). Calculate the acceleration of the mass. [Take \(g =10 m / s ^{2}\) ]
A. \(3.7 m / s ^{2}\) B. \(4.5 m / s ^{2} \) C. \(5.0 m / s ^{2}\) D. \(6.7 m / s ^{2}\)
Correct Answer: A
Explanation
By Newton's second law, \begin{aligned} &\vec{f}_{n e t}=\overrightarrow{m a} \\ &\vec{f}_{n e t}=F-f r=60 \cos 30-\mu m g \\ &=60 \cos 30-0.15 \times 10 \times 10=m a \\ &60(0.866)-15=10 a \\ &10 a=51.96-15 \\ &a=\frac{36.96}{10}=3.696 \\ &\therefore a \approx 3.7 m / s ^{2} \end{aligned}
