A 32.0 kg crate is initially moving with a velocity that has magnitude 3.58 m/s in a direction 37.0 ∘ west of north. How much work must be done on the crate to change its velocity to 5.10 m/s in a direction 63.0 ∘ south of east?

The amount of work done on the crate to change its velocity is 211.1 J.What is the work done on the crate to change its velocity?The work done on the crate to change its velocity is calculated by applying the principle of conservation of energy as shown below.Applying work-energy theorem, the work done in changing the velocity of the crate is equal to the change in kinetic energy of the crate as shown below;W = ΔK.EW = ¹/₂m(v₂² - v₁²)where;m is mass of the cratev₁ is the initial velocity of the cratev₂ is the final velocity of the crateW = ¹/₂m(v₂² - v₁²)W = ¹/₂(32)(5.1² - 3.58²)W = 211.1 JThus, the amount of work done on the crate to change its velocity from 3.58 m/s at 37.0 ∘ west of north to  5.10 m/s in a direction 63.0 ∘ south of east, is based on the principle of conservation of energy.Learn more about conservation of energy here: https://brainly.com/question/166559#SPJ1