Calculate the acceleration of the block and trolley system shown in the figure. The coefficient of kinetic fri

Question

Calculate the acceleration of the block and trolley system shown in the figure. The coefficient of kinetic friction between the trolley and the surface is $0.05$. (Take $g=10 	ext{ m/s}^2$, the mass of the string is negligible and no other friction exists). [Note: Based on the answer, the mass of the hanging block is $2 	ext{ kg}$ and the mass of the trolley is $10 	ext{ kg}$].

Accepted Answer

1. **Analyze the System:** The system consists of a trolley of mass $M$ (typically $10 	ext{ kg}$ in this specific PYQ) on a horizontal surface and a block of mass $m$ (typically $2 	ext{ kg}$) hanging vertically, connected by a string over a pulley.
2. **Identify Forces:**
   - **For the hanging block ($m$):** Gravity acts downwards ($mg$) and tension ($T$) acts upwards. Equation of motion: $mg - T = ma$.
   - **For the trolley ($M$):** Tension ($T$) acts forward and kinetic friction ($f_k$) acts backward. Equation of motion: $T - f_k = Ma$.
3. **Calculate Friction:** The normal reaction on the trolley is $N = Mg$. The kinetic friction is $f_k = \mu_k N = \mu_k Mg$ [Source 127, 134].
4. **Solve for Acceleration ($a$):** Adding the two equations eliminates tension $T$:
   $$ mg - f_k = (M + m)a $$
   $$ mg - \mu_k Mg = (M + m)a $$
   $$ a = \frac{g(m - \mu_k M)}{M + m} $$
   [Source 129]
5. **Calculation:** Substituting the standard values for this problem ($m=2 	ext{ kg}, M=10 	ext{ kg}, \mu_k=0.05, g=10 	ext{ m/s}^2$):
   $$ a = \frac{10(2 - 0.05 	imes 10)}{10 + 2} $$
   $$ a = \frac{10(2 - 0.5)}{12} = \frac{10(1.5)}{12} = \frac{15}{12} = 1.25 	ext{ m/s}^2 $$

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