Final answer:
The normal force of the slope acting on a 1.0 x 10^2 N sled held in place on a frictionless 20-degree slope by a rope is calculated using the equation N = W x cos(θ), which results in a normal force of approximately 93.97 N.
Step-by-step explanation:
The student's question pertains to the normal force of a slope acting on a sled when the sled is being held in place on a frictionless 20-degree slope by a rope. To find the normal force, we can use the concept that the normal force is perpendicular to the surface of the slope. In this case, we need to calculate the component of the sled's weight that is perpendicular to the slope since the sled is stationary and there is no motion perpendicular to the slope.
For a sled with weight W = m × g (mass multiplied by the acceleration due to gravity), the normal force N on a slope is calculated by multiplying the weight by the cosine of the slope's angle: N = W × cos(θ). If the weight of the sled is 1.0 x 102 N and the angle θ is 20 degrees, the normal force N would be calculated as follows:
N = (1.0 x 102 N) × cos(20°)
Plugging in the values and calculating using a calculator set to degree mode, we find the normal force to be:
N ≈ 1.0 x 102 N × 0.9397N ≈ 93.97 N
The normal force of the slope acting on the sled is approximately 93.97 N.