An object has a mass of 2 kg. The object's kinetic energy uniformly changes from 160 KJ to 36KJ over t in [0, 6 s]. What is the average speed of the object?

1 Answer
Narad T.
May 21, 2018

The average speed is =307.4ms^-1

Explanation:

The kinetic energy is

KE=1/2mv^2

The mass is m=2kg

The initial velocity is =u_1ms^-1

The final velocity is =u_2 ms^-1

The initial kinetic energy is 1/2m u_1^2=160000J

The final kinetic energy is 1/2m u_2^2=36000J

Therefore,

u_1^2=2/2*160000=160000m^2s^-2

and,

u_2^2=2/2*36000=36000m^2s^-2

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

The points are (0,160000) and (6, 36000)

The equation of the line is

v^2-160000=(36000-160000)/6t

v^2=-20666.7t+160000

So,

v=sqrt(-20666.7t+160000)

We need to calculate the average value of v over t in [0,6]

(6-0)bar v=int_0^6(sqrt(-20666.7t+160000))dt

6 barv= [(-20666.7t+160000)^(3/2)/(3/2*-20666.7)] _( 0) ^ (6)

=((-20666.7*6+160000)^(3/2)/(-31000))-((-20666.7*0+160000)^(3/2)/(-31000))

=160000^(3/2)/31000-36000^(3/2)/31000

=1844.2

So,

barv=1844.2/6=307.4ms^-1

The average speed is =307.4ms^-1

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