Two packing crates of masses 10.0 kg and 5.00 kg are connected by a light string that passes over a frictionless pulley. The 5.00-kg crate lies on a smooth incline of angle 40.0?. Find the acceleration of the 5.00-kg crate and the tension in the string.

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fudzisako · Accepted Answer

To solve this problem, we can use Newton&#039;s second law of motion, taking into account the forces acting on each crate.Let&#039;s denote:- m1=10.0 kg as the mass of the first crate- m2=5.00 kg as the mass of the second crate- a as the acceleration of the system- T as the tension in the string connecting the two cratesFor the 10.0-kg crate:The only force acting on this crate is its weight, which can be calculated as m1·g, where g is the acceleration due to gravity (approximately 9.8 m/s^2).Fweight1=m1·gFor the 5.00-kg crate:The force of gravity can be divided into two components:1. The component acting down the incline is m2·g·sin(θ).2. The component perpendicular to the incline is m2·g·cos(θ), which is balanced by the normal force exerted by the incline.The net force acting on the 5.00-kg crate is given by m2·g·sin(θ).Now, applying Newton&#039;s second law to each crate:For the 10.0-kg crate:Fnet1=m1·a=TFor the 5.00-kg crate:Fnet2=m2·a=m2·g·sin(θ)Since the crates are connected by a light string, the tension in the string is the same for both crates. Therefore, T=m2·g·cos(θ).Substituting the values and solving the equations, we have:m2·a=m2·g·sin(θ)a=g·sin(θ)T=m2·g·cos(θ)Substituting the given values:a=9.8·sin(40.0∘)T=5.00·9.8·cos(40.0∘)Evaluating these expressions, we find:a≈6.30 m/s² (rounded to two decimal places)T≈33.4 N (rounded to one decimal place)Therefore, the acceleration of the 5.00-kg crate is approximately 6.30 m/s², and the tension in the string is approximately 33.4 N.

Two packing crates of masses 10.0 kg and

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