A 495-kg dragster accelerates from rest to a final speed of 105 m/s in 395 m, during which it encounters an average frictional force of 1400 N. show answer No Attempt 50% Part (a) What is its average power output of the dragster, in watts, if this takes 8.2 s

Explanation:

According to energy conservation which states that the workdone is equal to change in the system

Workdone = change in kinetic energy + (frictional force * distance)

Workdone = ΔK + fd

Workdone = kf-Ki + fd

Workdone = = 1/2(m(v-u)^2) + fd

Given

Mass m = 495kg

final velocity v = 105m/s

initial velocity = 0m/s

Force f= 1400N

distance d = 395m

Substitute

Workdone = 1/2(495(105-0)^2) + 1400(395)

Workdone = 2,728,687.5+553000

Workdone = 3,281,687.5 Joules

Time = 8.2secs

Power output = Workdone/Time

Power output = 3,281,687.5/8.2

Power output = 885,766.768

Power output = 8.858 * 10^5 watts