- #1
janelle1905
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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
Wnet = 1/2mv2
v2 = v20 + 2(F/m)d
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!)