A 12 kg block is on an inclined plane at 30 degrees. The coefficient of friction is 2/3. Calculate the normal force and the component of weight perpendicular to the plane.

May 21, 2026Physics

ScanSolve AI Answer

May 21, 2026

Answer

12 \text{ kg}$

Here's the solution for question 8:

Given: Mass $m = 12 \text{ kg}$ Angle of inclination $\theta = 30^\circ$ Coefficient of friction $\mu = \frac{2}{3}$ Acceleration due to gravity $g = 10 m/s^2$

Step 1: Calculate the components of the weight. The weight of the body is $W = mg$ . $W = 12 kg \times 10 m/s^2 = 120 \text{ N}$ The component of weight perpendicular to the plane is $W_\perp = W \cos \theta$ .

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Here's the solution for question 8: Given: Mass m = 12 kg Angle of inclination = 30^ Coefficient of friction = (2)/(3) Acceleration due to gravity g = 10 m/s^2 Step 1: Calculate the components of the weight.

Asked on May 21, 2026Physics

This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.

ScanSolve AI AnswerAnswered on May 21, 2026

Here's the solution for question 8: Given: Mass m = 12 kg Angle of inclination = 30^ Coefficient of friction = (2)/(3) Acceleration due to gravity g = 10 m/s^2 Step 1: Calculate the components of the weight. The weight of the body is W = mg. W = 12 kg × 10 m/s^2 = 120 N The component of weight perpendicular to the plane is W_ = W . W_ = 120 \