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

[Ockonal]

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?

 

[Andrew Mason]

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

 

[Ockonal]

Ok, nevermind. Solved it, thanks.

 

[ashishsinghal]

A_res = m/2 × (V₂² - V₁²)

 

[rwisz]

I would like to clarify that the equations provided in the homework statement do not apply to this scenario. In order to change the speed of a body, work needs to be done by applying a force over a distance. The correct equation to calculate the work needed is W = Fd, where W is work, F is force, and d is distance. In this case, we do not have enough information to calculate the work needed, as we do not have the force or distance values.

Additionally, the equations provided in the "Attempt at a Solution" section do not apply to this situation as they involve potential energy, force, and kinetic energy, which are not relevant to the question asked. It is important to use the correct equations and understand the context in which they are applicable.

In summary, the work needed to change the speed of a body with a mass of 6×10⁴ kg from 2 m/s to 60 m/s cannot be calculated without knowing the force and distance values. Further information is needed to accurately answer this question.