Ask Question
2 September, 18:06

In physics, the kinetic energy of a moving object is given by the formula K=1/2mv^2 where m is the mass of the object and v is the object's velocity. Suppose a rocket is increasing in velocity at 10m/sec^2, and is decreasing in mass at 15kg/sec because it is using up fuel. How fast is it's total kinetic energy changing when the velocity is 30m/sec and the mass is 1000kg? Round to the nearest tenth.

+3
Answers (1)
  1. 2 September, 18:33
    0
    You need to take the derivative of both sides of the kinetic energy equation and then plug in the necessary values. The change in kinetic energy is dK/dt. If you take the derivative of the other side, you have to use product rule (since both m and v are variables). This gives dK/dt = (1/2) (dm/dt) (v^2) + (2) (1/2) mv (dv/dt)

    = (1/2) (dm/dt) (v^2) + mv (dv/dt), where dm/dt is the rate of change of mass, and dv/dt is the rate of change of the velocity at the given time.

    If you plug in for the given values (dm/dt = - 15, dv/dt = 10, m = 1000, v = 30), then you get dK/dt = (1/2) (-15) (30^2) + (1000) (30) (10) = - 6750 + 300,000 = 293,250 J/s. Notice that the answer is in units of Joules per second, beause you took the first derivative of kinetic energy with respect to time.
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “In physics, the kinetic energy of a moving object is given by the formula K=1/2mv^2 where m is the mass of the object and v is the object's ...” in 📗 Mathematics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers