How Much Work is Required to Accelerate a Mass from 2 m/s to 60 m/s?

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To calculate the work required to accelerate a mass of 6×10⁴ kg from 2 m/s to 60 m/s, the relevant equation is the change in kinetic energy, expressed as ΔKE = 0.5m(v₂² - v₁²). An initial attempt to calculate the work resulted in a negative value, indicating a misunderstanding in the application of the formula. The correct approach involves using A_res = m/2 × (V₂² - V₁²) to find the positive work done. The discussion highlights the importance of correctly applying kinetic energy principles in physics problems. The issue was resolved with the correct formula, leading to an accurate calculation of work.
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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 = mv²/2


The 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?
 
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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 = mv²/2

The 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?
The change in kinetic energy is positive: \Delta KE = \Delta (mv^2/2) = .5m(v_2^2-v_1^2)

What units does your answer have? Why do you think the result is too big?

AM
 
Ok, nevermind. Solved it, thanks.
 
A_res = m/2 × (V₂² - V₁²)
 
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