# Homework Help: The Work-Energy Principle & Kinematic Eq'ns to calculate speed

1. Aug 31, 2009

### janelle1905

1. The problem statement, all variables and given/known data

A spaceship of mass 5.00 x 104 kg is travelling 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.

2. Relevant equations

Wnet = 1/2mv2
v2 = v20 + 2(F/m)d

3. The attempt at a solution

a. Using work energy theorem:
Wnet = W + Wfr = Fdcos0o + 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!)

2. Aug 31, 2009

### kuruman

You misquoted the work-energy theorem. It is

Wnet = ΔK

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

3. Aug 31, 2009

### janelle1905

Okay...So in my Wnet calculation, should it be Wnet = 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?

4. Aug 31, 2009

### 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(vfinal)2 - (1/2)m(vinitial)2 = Wnet

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

You need to use the above equation with vinitial = 1.15x104 m/s to find vfinal. 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 in to check your numbers.

5. Aug 31, 2009

### 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!