A tractor pulls a sled of mass \(M=1000\). kg across level ground. The coefficient of kinetic friction between the sled and the ground is \(\mu_{k}=0.600 .\) The tractor pulls the sled by a rope that connects to the sled at an angle of \(\theta=30.0^{\circ}\) above the horizontal. What magnitude of tension in the rope is necessary to move the sled horizontally with an acceleration \(a=2.00 \mathrm{~m} / \mathrm{s}^{2} ?\)

Draw a free body diagram for the sled, indicating all the forces acting on it. The forces include the gravitational force (M × g), the normal force (Fn), the tension force in the rope (T), and the kinetic friction force (Fk).

The tension force can be resolved into a horizontal component (Tx) and a vertical component (Ty) as follows: Tx = T × cos(θ) Ty = T × sin(θ)

In the vertical direction, the sled is in equilibrium. The following equation can be written: Fn - Ty - M × g = 0

In the horizontal direction, the following equation can be written using Newton's second law of motion: M × a = Tx - Fk

The kinetic friction force can be written as: Fk = μk × Fn

Plug in the values for M, g, μk, θ, and a into the previous equations and solve for T. Use the following values: M = 1000 kg, g = 9.81 m/s², μk = 0.600, θ = 30°, and a = 2.00 m/s². First, find Fn using the vertical forces equation: Fn = M × g + Ty Fn = M × g + T × sin(θ) Now, find Fk using the kinetic friction equation: Fk = μk × Fn Substitute Fn and Fk into the horizontal forces equation: M × a = Tx - Fk M × a = T × cos(θ) - μk × Fn Substitute Fn equation in the above equation and solve for T: T = M × a / (cos(θ) - μk × sin(θ)) Finally, substitute the given values and solve for T: T = 1000 × 2 / (cos(30°) - 0.600 × sin(30°)) T ≈ 2395.36 N Thus, the magnitude of tension in the rope necessary to move the sled horizontally with an acceleration of 2.00 m/s² is approximately 2395.36 N.