Question

A block of metal, of $25\text{ g}$ mass moves down without acceleration when the plane is inclined at an angle of $30^\circ$. When the inclination is increased by $30^\circ$, find the downward acceleration of the block.

• A. $1.9\text{ m s}^{-2}$
• B. $2.6\text{ m s}^{-2}$
• C. $5.66\text{ m s}^{-2}$
• D. $3.8\text{ m s}^{-2}$

Hint:

For any incline problem where a block slides at constant velocity at $30^\circ$ and the angle is then increased to $60^\circ$, the net acceleration expression always simplifies beautifully to $a = \frac{g}{\sqrt{3}}$. Memorizing this specific geometry shortcut saves immense algebraic effort during an exam!

Answer 1

The Correct Option is C

Concept: When a block moves down an inclined plane without acceleration, it is moving with a constant velocity. This means the net force acting along the incline surface is zero, indicating that the downward component of gravity is perfectly balanced by the upward kinetic friction force ($mg \sin\theta = f_k = \mu_k mg \cos\theta$). This gives: \[ \mu_k = \tan\theta \] When the angle of inclination is increased, a net driving force is established, and the resulting acceleration is given by Newton's second law: \[ a = g(\sin\theta' - \mu_k \cos\theta') \]

Step 1: Determine the coefficient of kinetic friction ($\mu_k$).
Given that the block slides at constant speed when $\theta = 30^\circ$: \[ \mu_k = \tan(30^\circ) = \frac{1}{\sqrt{3}} \]

Step 2: Calculate the acceleration at the new inclination angle.
The inclination is increased by $30^\circ$, making the new angle: \[ \theta' = 30^\circ + 30^\circ = 60^\circ \] Substitute $\theta' = 60^\circ$, $\mu_k = \frac{1}{\sqrt{3}}$, and $g = 9.8\text{ m s}^{-2}$ into the acceleration equation: \[ a = g\left(\sin(60^\circ) - \frac{1}{\sqrt{3}}\cos(60^\circ)\right) \] We know that $\sin(60^\circ) = \frac{\sqrt{3}}{2}$ and $\cos(60^\circ) = \frac{1}{2}$: \[ a = g\left(\frac{\sqrt{3}}{2} - \frac{1}{\sqrt{3}}\cdot\frac{1}{2}\right) = g\left(\frac{\sqrt{3}}{2} - \frac{1}{2\sqrt{3}}\right) \] Find a common denominator inside the parentheses: \[ a = g\left(\frac{3 - 1}{2\sqrt{3}}\right) = g\left(\frac{2}{2\sqrt{3}}\right) = \frac{g}{\sqrt{3}} \] Substituting $g = 9.8\text{ m s}^{-2}$ and $\sqrt{3} \approx 1.732$: \[ a = \frac{9.8}{1.732} \approx 5.658\text{ m s}^{-2} \approx 5.66\text{ m s}^{-2} \]