A 1600 kg vehicle moves with a velocity of 19.5 m/s. Calculate the power required to reduce the velocity to 3.20 m/s in 11.0 s.
I don't know this one the only thing I know is p=w/t
dont know if its correct but its something.
p=w/t
W=KE=(1/2)m(v)^2
p=(.5mv^2)/t
p_inital=(.5(1600)(19.5))/11=27654watts
p_final=(.5(1600)(3.2))/11=744.7...watts
(p_final)-(p_inital)=26909.27...watts
maybe? idk
To calculate the power required to reduce the velocity of the vehicle, you can use the work-energy principle, which states that the work done on an object is equal to the change in its kinetic energy. The power can then be calculated by dividing the work by the time taken to perform that work.
First, let's calculate the initial kinetic energy (KEi) of the vehicle using the formula: KEi = 0.5 * mass * velocity^2.
KEi = 0.5 * 1600 kg * (19.5 m/s)^2
KEi = 0.5 * 1600 kg * 380.25 m^2/s^2
KEi = 304,200 J (Joules)
Next, calculate the final kinetic energy (KEf) of the vehicle using the same formula, but with the final velocity:
KEf = 0.5 * 1600 kg * (3.20 m/s)^2
KEf = 0.5 * 1600 kg * 10.24 m^2/s^2
KEf = 8,192 J (Joules)
Now, calculate the work done (W) by subtracting the final kinetic energy from the initial kinetic energy:
W = KEi - KEf
W = 304,200 J - 8,192 J
W = 296,008 J (Joules)
Finally, calculate the power (P) required to do this work by dividing the work by the time taken (t):
P = W / t
P = 296,008 J / 11.0 s
P ≈ 26,910.73 W (Watts)
Therefore, the power required to reduce the velocity of the vehicle from 19.5 m/s to 3.20 m/s in 11.0 s is approximately 26,910.73 Watts.