3. A length of rope is used to pull a mass $m$ block along a horizontal surface as shown. The tension $\vec{T}$. in the rope is applied at an angle $\phi$ above the horizontal, the horizontal surface has coefficient of kinetic friction $\mu_{k}$ with the block as it slides, and the block has acceleration $\vec{a}$. Which expression gives the magnitude of the tension, T? Assume the acceleration due to gravity is $a_{y}=-g$.
A. $\frac{m a}{\cos \phi}$
B. $\frac{\mu_{k} m a}{\tan \phi}$
C. $\frac{m\left(a+\mu_{k} g\right)}{\cos \phi+\mu_{k} \sin \phi}$
D. $\frac{m a}{\cos \phi-\mu_{k} \sin \phi}$.
Answer
Combine the terms to get the final expression for T: \(\boxed{T = \frac{m(a + \mu_k g)}{\cos(\phi) + \mu_k\sin(\phi)}}\)
Step 1 ：Break the tension T into its horizontal and vertical components: \(T_x = T\cos(\phi)\) and \(T_y = T\sin(\phi)\)
Step 2 ：Write Newton's second law equations for horizontal and vertical forces: \(-N\mu_k + T\cos(\phi) = ma\) and \(N + T\sin(\phi) - mg = 0\)
Step 3 ：Solve the system of equations for T: \(T = \frac{ma}{\cos(\phi) + \mu_k\sin(\phi)} + \frac{mg\mu_k}{\cos(\phi) + \mu_k\sin(\phi)}\)
Step 4 ：Combine the terms to get the final expression for T: \(\boxed{T = \frac{m(a + \mu_k g)}{\cos(\phi) + \mu_k\sin(\phi)}}\)
