Final answer:
The normal force exerted by the incline on a 2.00 kg block in equilibrium at an angle of 36.0° is approximately 15.88 N, which would be rounded to the nearest given option of 16.0 N (assuming a typo in options).
This correct answer is none of the above.
Step-by-step explanation:
The student asked for the force of the incline on a 2.00 kg block that is in equilibrium on an incline of angle 36.0°. To find the force exerted by the incline on the block, also known as the normal force (N), we can use the concept of equilibrium where the sum of forces in all directions is zero.
Since there is no motion perpendicular to the incline, the normal force must balance the component of gravitational force acting perpendicular to the incline.
The gravitational force acting on the block can be calculated as weight (W) = mass (m) × gravity (g), where g = 9.8 m/s². So, for a block of mass 2.00 kg,
W = 2.00 kg × 9.8 m/s² = 19.6 N. This force acts directly downwards. However, we are interested in the component of this force that acts perpendicular to the inclined plane. This component is given by W·cos(θ), where θ is the angle of the incline.
Thus, N = W·cos(36.0°) = 19.6 N × cos(36.0°) ≈ 19.6 N × 0.809 = 15.88 N, which rounds to the nearest option which is (b) 16.0 N, assuming it is a typo because none of the provided options are exact. The approximation arises due to the use of a rounded value for the cosine of the angle.
This correct answer is none of the above.