Question:
A $5\text{ kg}$ block on a horizontal surface is pulled by a force of $15\text{ N}$. If the coefficient of friction between the block and the surface is $0.2$, then the acceleration of the block is [Take $g = 10\text{ ms}^{-2}$]:
A $5\text{ kg}$ block on a horizontal surface is pulled by a force of $15\text{ N}$. If the coefficient of friction between the block and the surface is $0.2$, then the acceleration of the block is [Take $g = 10\text{ ms}^{-2}$]:
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Always calculate the maximum limiting friction ($f = \mu mg$) first. If the applied force were less than $10\text{ N}$ (for example, $8\text{ N}$), the acceleration would be $0\text{ ms}^{-2}$, because static friction matches the applied force up to its limiting threshold.
Updated On: Jun 10, 2026
- $0\text{ ms}^{-2}$
- $1\text{ ms}^{-2}$
- $2\text{ ms}^{-2}$
- $3\text{ ms}^{-2}$
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The Correct Option is B
Solution and Explanation
Concept:
When a force is applied to a body resting on a rough surface, the motion is opposed by friction. To determine if the body moves and to find its acceleration, we must compare the applied force ($F_{\text{applied}}$) with the maximum limiting static friction force ($f_{s,\max} = \mu N$).
• If $F_{\text{applied}} \le f_{s,\max}$, the block remains stationary ($a = 0$), and the static friction balances the applied force.
• If $F_{\text{applied}} > f_{s,\max}$, the block moves, and the net accelerating force is $F_{\text{net}} = F_{\text{applied}} - f_k$.
Step 1: Let us sketch the forces operating along the vertical axis of motion. Since there is no vertical movement, the normal reaction force ($N$) provided by the horizontal platform perfectly balances the gravitational weight ($mg$) of the block: $$N = mg$$ Given values: $$\text{Mass } m = 5\text{ kg}$$ $$\text{Acceleration due to gravity } g = 10\text{ ms}^{-2}$$ Substituting these parameters yields: $$N = 5\text{ kg} \times 10\text{ ms}^{-2} = 50\text{ N}$$
Step 2: The maximum threshold value of frictional resistance that the contact surface can produce is defined by: $$f_{\max} = \mu \cdot N$$ Given that the coefficient of friction $\mu = 0.2$: $$f_{\max} = 0.2 \times 50\text{ N} = 10\text{ N}$$
Step 3: The applied horizontal pulling force is explicitly given as: $$F_{\text{applied}} = 15\text{ N}$$ Comparing these two values: $$15\text{ N} > 10\text{ N} \implies F_{\text{applied}} > f_{\max}$$ Since the applied force exceeds the limiting friction threshold, the block will overcome static friction and accelerate across the surface.
Step 4: The net force acting along the horizontal direction of motion is the difference between the pulling force and the opposing frictional force: $$F_{\text{net}} = F_{\text{applied}} - f_{\max}$$ $$F_{\text{net}} = 15\text{ N} - 10\text{ N} = 5\text{ N}$$ According to Newton's Second Law ($F_{\text{net}} = m \cdot a$), the resulting acceleration $a$ is: $$a = \frac{F_{\text{net}}}{m}$$ $$a = \frac{5\text{ N}}{5\text{ kg}} = 1\text{ ms}^{-2}$$ Thus, the block accelerates at a rate of $1\text{ ms}^{-2}$, matching Option (B).
• If $F_{\text{applied}} \le f_{s,\max}$, the block remains stationary ($a = 0$), and the static friction balances the applied force.
• If $F_{\text{applied}} > f_{s,\max}$, the block moves, and the net accelerating force is $F_{\text{net}} = F_{\text{applied}} - f_k$.
Step 1: Let us sketch the forces operating along the vertical axis of motion. Since there is no vertical movement, the normal reaction force ($N$) provided by the horizontal platform perfectly balances the gravitational weight ($mg$) of the block: $$N = mg$$ Given values: $$\text{Mass } m = 5\text{ kg}$$ $$\text{Acceleration due to gravity } g = 10\text{ ms}^{-2}$$ Substituting these parameters yields: $$N = 5\text{ kg} \times 10\text{ ms}^{-2} = 50\text{ N}$$
Step 2: The maximum threshold value of frictional resistance that the contact surface can produce is defined by: $$f_{\max} = \mu \cdot N$$ Given that the coefficient of friction $\mu = 0.2$: $$f_{\max} = 0.2 \times 50\text{ N} = 10\text{ N}$$
Step 3: The applied horizontal pulling force is explicitly given as: $$F_{\text{applied}} = 15\text{ N}$$ Comparing these two values: $$15\text{ N} > 10\text{ N} \implies F_{\text{applied}} > f_{\max}$$ Since the applied force exceeds the limiting friction threshold, the block will overcome static friction and accelerate across the surface.
Step 4: The net force acting along the horizontal direction of motion is the difference between the pulling force and the opposing frictional force: $$F_{\text{net}} = F_{\text{applied}} - f_{\max}$$ $$F_{\text{net}} = 15\text{ N} - 10\text{ N} = 5\text{ N}$$ According to Newton's Second Law ($F_{\text{net}} = m \cdot a$), the resulting acceleration $a$ is: $$a = \frac{F_{\text{net}}}{m}$$ $$a = \frac{5\text{ N}}{5\text{ kg}} = 1\text{ ms}^{-2}$$ Thus, the block accelerates at a rate of $1\text{ ms}^{-2}$, matching Option (B).
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