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. $0\text{ ms}^{-2}$
• B. $1\text{ ms}^{-2}$
• C. $2\text{ ms}^{-2}$
• D. $3\text{ ms}^{-2}$

Hint:

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.

Answer 1

The Correct Option is B

Step 1: List the given values.
The block has mass $m = 5\text{ kg}$. The pulling force is $F = 15\text{ N}$. The coefficient of friction is $\mu = 0.2$, and we take $g = 10\text{ ms}^{-2}$.

Step 2: Find the normal force.
The surface is horizontal, so the normal force equals the weight of the block: \[ N = m g = 5 \times 10 = 50\text{ N} \]

Step 3: Find the friction force.
Friction opposes the motion and its size is: \[ f = \mu N = 0.2 \times 50 = 10\text{ N} \] This is the resistance the block feels as it slides.

Step 4: Check that the block actually moves.
The applied force is $15\text{ N}$ and the friction is $10\text{ N}$. Since $15 > 10$, the pull wins and the block slides forward.

Step 5: Find the net force.
Subtract friction from the applied force: \[ F_{net} = 15 - 10 = 5\text{ N} \]

Step 6: Use Newton's second law.
Acceleration equals net force divided by mass: \[ a = \frac{F_{net}}{m} = \frac{5}{5} = 1\text{ ms}^{-2} \] So the block speeds up at one metre per second squared. \[ \boxed{1\text{ ms}^{-2}} \]