The discussion revolves around calculating the work required to accelerate a mass of 6×10⁴ kg from an initial speed of 2 m/s to a final speed of 60 m/s, focusing on the concepts of work and kinetic energy.
The conversation includes attempts to clarify the calculations and units involved. One participant expresses confusion about the negative result, while another suggests a correction in the formula used for calculating work. A participant indicates they have resolved their issue, but the details of that resolution are not shared.
There is an emphasis on understanding the relationship between kinetic energy and work, with participants questioning the assumptions made in their calculations. The original poster's approach and the subsequent discussions highlight potential misunderstandings in applying the relevant equations.
The change in kinetic energy is positive: \Delta KE = \Delta (mv^2/2) = .5m(v_2^2-v_1^2)Ockonal said:Homework Statement
What work needs to be done to change body speed with mass 6×10⁴ kg from 2 m/s to 60 m/s?Homework Equations
A = mgh
A = FS
A = mv²/2The Attempt at a Solution
A₁ = mV₁² / 2
A₂ = mV₂² / 2
A_res = A₂ - A₁;
A_res = m/2 × (V₁² - V₂²)
A_res = 3×10⁴ × (4 - 3600)
The result is negative and too big. Is that right?