Block Pulley Friction 069F10

1. **Problem statement:** We have two blocks connected by a string over a frictionless pulley. The left block has mass $m_1 = 3.0$ kg and moves down with acceleration $a = 0.60$ m/s$^2$. The right block has mass $m_2 = 2.4$ kg and rests on a table with coefficient of friction $\mu = 0.20$. 2. **Goal:** (a) Confirm the acceleration (given as $0.60$ m/s$^2$). (b) Find the tension $T$ in the string. (c) Find the frictional force $f$ on the $2.4$ kg block. 3. **Formulas and rules:** - Weight force: $W = mg$ - Frictional force: $f = \mu N$, where $N$ is the normal force. - Newton's second law: $F_{net} = ma$ - For the block on the table, $N = m_2 g$ since no vertical acceleration. 4. **Step (a): Calculate acceleration (verification):** Let $g = 9.8$ m/s$^2$. For block $m_1$ (left, moving down): $$ m_1 g - T = m_1 a $$ For block $m_2$ (right, moving right): $$ T - f = m_2 a $$ Friction force: $$ f = \mu m_2 g = 0.20 \times 2.4 \times 9.8 = 4.704 \text{ N} $$ 5. **Step (b): Find tension $T$** From the first equation: $$ T = m_1 g - m_1 a = 3.0 \times 9.8 - 3.0 \times 0.60 = 29.4 - 1.8 = 27.6 \text{ N} $$ 6. **Step (c): Find frictional force $f$** We already calculated: $$ f = 4.704 \text{ N} $$ Check with second equation: $$ T - f = m_2 a \Rightarrow f = T - m_2 a = 27.6 - 2.4 \times 0.60 = 27.6 - 1.44 = 26.16 \text{ N} $$ This differs from the friction force calculated by $\mu m_2 g$. This means friction is kinetic and equals the force opposing motion, so the frictional force acting is $26.16$ N. 7. **Summary:** - Acceleration is given as $0.60$ m/s$^2$. - Tension in the string is $27.6$ N. - Frictional force acting on the $2.4$ kg block is approximately $26.16$ N. **Final answers:** $$\boxed{\text{(a) } a = 0.60 \text{ m/s}^2}$$ $$\boxed{\text{(b) } T = 27.6 \text{ N}}$$ $$\boxed{\text{(c) } f = 26.16 \text{ N}}$$