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Question

A block of mass $5 \text{ kg}$ is moving on an inclined plane which makes an angle of $30^\circ$ with the horizontal. Friction coefficient between the block and inclined plane surface is $\frac{\sqrt{3}}{2}$. The force to be applied on the block so that the block will move down without acceleration is ________$\text{N}$.
($g = 10 \text{ m/s}^2$).

The correct answer is
$7.5$

Force Calculation for Zero Acceleration on Inclined Plane

To find the force required for the block to move down the inclined plane without acceleration ($a=0$), we analyze the forces acting parallel to the plane and set the net force to zero.

Forces Acting Parallel to the Incline

  • Component of Gravity: Acts downwards. Calculated as $W_\parallel = mg \sin\theta$.
  • Kinetic Friction Force: Acts upwards (opposing downward motion). Calculated as $f_k = \mu_k N$, where $N = mg \cos\theta$.
  • Applied Force: External force $F$, acting parallel to the incline.

Step-by-Step Calculation

  1. Given Values:
    • Mass, $m = 5 \text{ kg}$
    • Angle, $\theta = 30^\circ$
    • Friction coefficient, $\mu_k = \frac{\sqrt{3}}{2}$
    • Gravity, $g = 10 \text{ m/s}^2$
  2. Calculate Downward Gravity Component:

    $W_\parallel = mg \sin\theta = (5 \text{ kg})(10 \text{ m/s}^2) \sin(30^\circ) = 50 \times \frac{1}{2} = 25 \text{ N}$

  3. Calculate Normal Force and Friction:

    Normal Force, $N = mg \cos\theta = (5 \text{ kg})(10 \text{ m/s}^2) \cos(30^\circ) = 50 \times \frac{\sqrt{3}}{2} = 25\sqrt{3} \text{ N}$

    Friction Force, $f_k = \mu_k N = \left(\frac{\sqrt{3}}{2}\right) (25\sqrt{3} \text{ N}) = \frac{25 \times 3}{2} = \frac{75}{2} = 37.5 \text{ N}$

  4. Net Force Equation for Zero Acceleration:

    Assuming the applied force $F$ acts downwards along the incline, the net force $F_{net}$ along the incline is the sum of forces acting downwards minus forces acting upwards.

    $F_{net} = W_\parallel + F - f_k$

    For zero acceleration, $F_{net} = 0$:

    $25 \text{ N} + F - 37.5 \text{ N} = 0$

  5. Solve for Applied Force $F$:

    $F = 37.5 \text{ N} - 25 \text{ N}$

    $F = 12.5 \text{ N}$

    The required applied force is $12.5 \text{ N}$ downwards along the incline.

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