A30.0-kg crate is initially moving with a velocity that has magnitude 3.90 m/s in a direction 37.0 degrees west of north. how much work must be done on the crate to change its velocity to 5.62 m/s in a direction 63.0 degrees south of east?

Answerthe energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.explanationwe define electric potential as the ratio electric potential energy to the charge,    then the following gives us the potential difference,      the formula  may be on your formula sheet.the change in electric potential energy would be the energy that would needed to move it from point a to point b. solving for that, we get    using q = 8.00  × 10⁻¹⁹ c and  δv = 2.00 v, we get    the energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.