Blaine skis down a hill at an angle of 12°. He has a mass of 82 kg. What is the normal force on Blaine?

The answer is . . . 786.8 N (see work attached)
normal force = (mg)cos(θ) = (82kg)(9.81 m/s²)cos(12⁰) = 786.8 N

Answered by Anonymous | 2024-06-10

Normal force on an incline is N = m ∗ g ∗ cos ( x ) Where m = 82 k g and g = 9.8 m / s 2
Therefore:
N = 82 ∗ 9.8 ∗ cos ( 12 ) N = 803.6 ∗ cos ( 12 ) N = 803.6 ∗ .98 N = 787.528
Normal force is 787.528 N.

Answered by ChrisR | 2024-06-10

The normal force acting on Blaine while skiing down a hill at an angle of 12° is approximately 786.8 N. This is calculated by finding the weight of Blaine and then taking the component of that weight that acts perpendicular to the slope. The formula used is N = mg cos(θ).
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Answered by Anonymous | 2024-11-01