Asked by Help
A hovercraft of mass 69.0 kg can move on a horizontal surface, the x-y plane. A single unbalanced force acts on the hovercraft, but the size of the force is unknown. The hovercraft initially has a velocity of 5.0 m/s in the positive x direction and some time later has a velocity of 2.0 m/s in the positive y direction. How much work is done on the hovercraft by the force during this time?
I'm having trouble visualizing and applying the appropriate work formula.
I'm having trouble visualizing and applying the appropriate work formula.
Answers
Answered by
drwls
The work done is what increases its kinetic energy, and equals the amount of that increase. (In this case, it is a decrease, so negative work is done). The directions are not important. Just compute the change in kinetic energy.
W = delta(KE) = -(1/2)(69)(25 - 4)
= -724.5 J
W = delta(KE) = -(1/2)(69)(25 - 4)
= -724.5 J