Question:
A \(3\,\text{kg}\) block on a rough incline of angle \(37^\circ\) is connected to a hanging mass of \(4\,\text{kg}\). If the coefficient of friction between the \(3\,\text{kg}\) block and the rough incline is \(\mu=0.25\), then the acceleration of the system is
\[
g=10\,\text{m s}^{-2},
\qquad
\sin37^\circ=0.6,
\qquad
\cos37^\circ=0.8
\]
A \(3\,\text{kg}\) block on a rough incline of angle \(37^\circ\) is connected to a hanging mass of \(4\,\text{kg}\). If the coefficient of friction between the \(3\,\text{kg}\) block and the rough incline is \(\mu=0.25\), then the acceleration of the system is
\[
g=10\,\text{m s}^{-2},
\qquad
\sin37^\circ=0.6,
\qquad
\cos37^\circ=0.8
\]
Show Hint
For connected body problems,
\[
a=\frac{\text{Driving Force}-\text{Resisting Force}}
{\text{Total Mass}}.
\]
Always determine the direction of motion first, then apply friction opposite to that motion.
Updated On: Jul 9, 2026
- \(2.28\,\text{m s}^{-2}\)
- \(1.08\,\text{m s}^{-2}\)
- \(3.2\,\text{m s}^{-2}\)
Zero
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The Correct Option is A
Solution and Explanation
Concept:
Since the hanging mass is larger, the \(4\,\text{kg}\) block moves downward and the \(3\,\text{kg}\) block moves upward along the incline.
Friction opposes the motion of the \(3\,\text{kg}\) block.
Step 1: Calculate the component of weight along the incline. For the \(3\,\text{kg}\) block, \[ W_{\parallel} = mg\sin37^\circ = 3\times10\times0.6 = 18\,\text{N}. \]
Step 2: Calculate the frictional force. Normal reaction, \[ N = mg\cos37^\circ = 3\times10\times0.8 = 24\,\text{N}. \] Hence, \[ f = \mu N = 0.25\times24 = 6\,\text{N}. \]
Step 3: Find the net driving force. Weight of the hanging block, \[ W=4g=40\,\text{N}. \] Opposing forces on the inclined block, \[ 18+6=24\,\text{N}. \] Therefore, \[ F_{\text{net}} = 40-24 = 16\,\text{N}. \]
Step 4: Calculate the acceleration. Total mass of the system, \[ M = 3+4 = 7\,\text{kg}. \] Hence, \[ a = \frac{F_{\text{net}}}{M} = \frac{16}{7} = 2.286\,\text{m s}^{-2}. \] \[ a \approx 2.28\,\text{m s}^{-2}. \] Final Answer: \[ \boxed{a=2.28\,\text{m s}^{-2}} \] \[ \boxed{\text{Answer = (A)}} \]
Step 1: Calculate the component of weight along the incline. For the \(3\,\text{kg}\) block, \[ W_{\parallel} = mg\sin37^\circ = 3\times10\times0.6 = 18\,\text{N}. \]
Step 2: Calculate the frictional force. Normal reaction, \[ N = mg\cos37^\circ = 3\times10\times0.8 = 24\,\text{N}. \] Hence, \[ f = \mu N = 0.25\times24 = 6\,\text{N}. \]
Step 3: Find the net driving force. Weight of the hanging block, \[ W=4g=40\,\text{N}. \] Opposing forces on the inclined block, \[ 18+6=24\,\text{N}. \] Therefore, \[ F_{\text{net}} = 40-24 = 16\,\text{N}. \]
Step 4: Calculate the acceleration. Total mass of the system, \[ M = 3+4 = 7\,\text{kg}. \] Hence, \[ a = \frac{F_{\text{net}}}{M} = \frac{16}{7} = 2.286\,\text{m s}^{-2}. \] \[ a \approx 2.28\,\text{m s}^{-2}. \] Final Answer: \[ \boxed{a=2.28\,\text{m s}^{-2}} \] \[ \boxed{\text{Answer = (A)}} \]
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