An object has a mass of 12 kg. The object's kinetic energy uniformly changes from 64 KJ to 24 KJ over t in [0, 5 s]. What is the average speed of the object?

1 Answer
Narad T.
Jun 18, 2017

The average speed is =84.9ms^-1

Explanation:

The kinetic energy is

KE=1/2mv^2

The mass is =12kg

The initial velocity is =u_1

The initial kinetic energy is 1/2m u_1^2=64000J

The final velocity is =u_2

The final kinetic energy is 1/2m u_2^2=24000J

Therefore,

u_1^2=2/12*64000=10666.7m^2s^-2

and,

u_2^2=2/12*24000=4000m^2s^-2

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

The points are (0,10666.7) and (5,4000)

The equation of the line is

v^2-10666.7=(4000-10666.7)/5t

v^2=-1333.3t+10666.7

So,

v=sqrt((-1333.3t+10666.7)

We need to calculate the average value of v over t in [0,5]

(5-0)bar v=int_0^5sqrt((-1333.3t+10666.7))dt

5 barv=[((-1333.3t+10666.7)^(3/2)/(-3/2*1333.3)]_0^5

=((-1333.3*5+10666.7)^(3/2)/(2000))-((-1333.3*0+10666.7)^(3/2)/(2000))

=10666.7^(3/2)/2000-4000^(3/2)/2000

=424.3

So,

barv=424.3/5=84.9ms^-1

The average speed is =84.9ms^-1

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