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
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.
1 answer
2 answers