An object has a mass of 4 kg4kg. The object's kinetic energy uniformly changes from 144 KJ144KJ to 120KJ120KJ over t in [0, 2 s]t[0,2s]. What is the average speed of the object?

1 Answer
Aug 14, 2017

The average speed is =256.8ms^-1=256.8ms1

Explanation:

The kinetic energy is

KE=1/2mv^2KE=12mv2

The mass is =4kg=4kg

The initial velocity is =u_1ms^-1=u1ms1

The final velocity is =u_2 ms^-1=u2ms1

The initial kinetic energy is 1/2m u_1^2=144000J12mu21=144000J

The final kinetic energy is 1/2m u_2^2=120000J12mu22=120000J

Therefore,

u_1^2=2/4*144000=72000m^2s^-2u21=24144000=72000m2s2

and,

u_2^2=2/4*120000=60000m^2s^-2u22=24120000=60000m2s2

The graph of v^2=f(t)v2=f(t) is a straight line

The points are (0,72000)(0,72000) and (2,60000)(2,60000)

The equation of the line is

v^2-72000=(60000-72000)/2tv272000=60000720002t

v^2=-6000t+72000v2=6000t+72000

So,

v=sqrt((-6000t+72000)v=(6000t+72000)

We need to calculate the average value of vv over t in [0,2]t[0,2]

(2-0)bar v=int_0^2(sqrt(-6000t+72000))dt(20)¯v=20(6000t+72000)dt

2 barv=[((-6000t+72000)^(3/2)/(3/2*-6000))]_0^22¯v=(6000t+72000)3232600020

=((-6000*2+72000)^(3/2)/(-9000))-((-6000*0+72000)^(3/2)/(-9000))=(60002+72000)329000(60000+72000)329000

=72000^(3/2)/9000-60000^(3/2)/9000=7200032900060000329000

=513.6=513.6

So,

barv=513.6/2=256.8ms^-1¯v=513.62=256.8ms1

The average speed is =256.8ms^-1=256.8ms1