The Work-Energy Principle & Kinematic Eq'ns to calculate speed

[janelle1905]

Homework Statement

A spaceship of mass 5.00 x 10⁴ kg is traveling at a speed 1.15 x 10⁴ m/s in outer space. Except for the force generated by its own engine, no other force acts on the ship. As the engine exerts a constant force of 4.00 x 10⁵ N, the ship moves a distance of 2.50 x 10⁶m in the direction of the force of the engine.
a. Determine the final speed of the ship using the work-energy theorem.
b. Determine the final speed of the ship using kinematic equations.

Homework Equations

W_net = 1/2mv²
v² = v²₀ + 2(F/m)d

The Attempt at a Solution

a. Using work energy theorem:
W_net = W + W_fr = Fdcos0^o + 0 = (4.00 x 10⁵)(2.50 x 10⁶) = 1.00 x 10¹²
v² = (2)(1.00x10¹²)/5.00x10⁴ = 6324.6 m/s

b. Using kinematic equations:
v² = (1.15 x 10⁴)² + 2(4.00 x 10⁵/5.00 x 10⁴)(2.50 x 10⁶)
v = 13,124 m/s

According to my calculation in part a, the final speed is 6324.6 m/s, however in part b my calculation shows that the velocity is 13,124 m/s. The difference between the two calcuations is the v₀ is included in the second one, but not in the first. It seems to me that v₀ should be included, but I don't know how to incorporate it into the work-energy theorem.

Thanks in advance for you help :)

(I know these answers aren't the correct number of sig figs yet...just trying to get the right answer first!)

 

[kuruman]

You misquoted the work-energy theorem. It is

W_net = ΔK

where ΔK is the change in kinetic energy (final minus initial). This spaceship is already moving when in fires its engine.

 

[janelle1905]

Okay...So in my Wnet calculation, should it be W_net = Fdcos0 + v⁰

I know that KE = 1/2mv², this is what the second term of the previous equation should be, however this leads to the wrong answer doesn't it?

Also, do you think my calculation of velocity using kinematic equations is correct?

Thanks for your help.

 

[kuruman]

You are missing the point. I am not talking about kinetic energy KE. I am talking about change in kinetic energy. The work-energy theorem in this case should be written as

(1/2)m(v_final)² - (1/2)m(v_initial)² = W_net

The term on the left is the change in kinetic energy, final minus initial.

You need to use the above equation with v_initial = 1.15x10⁴ m/s to find v_final. The work in this case is indeed Fd. I don't know what v⁰ means, but it should not be there.

The kinematic equation is correctly set up. I did not plug into check your numbers.

 

[janelle1905]

Okay - I understand it now. I plugged in and got the same answer for both part a and b this time.
Thanks very much for your help!