An object has a mass of 12 kg. The object's kinetic energy uniformly changes from 64 KJ to 140 KJ over t in [0, 5 s]. What is the average speed of the object?

1 Answer
Narad T.
Mar 13, 2017

The average speed is =129.6ms^-1

Explanation:

The kinetis energy is

KE=1/2mv^2

The initial velocity is =u_1

1/2m u_1^2=64000J

The final velocity is =u_2

1/2m u_2^2=140000J

Therefore,

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

and,

u_2^2=2/12*140000=23333.3m^2s^-2

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

The points are (0,1066.7) and (5,23333.3)

The equation of the line is

v^2-10666.7=(23333.3-10666.7)/5t

v^2=(12666.6)/5t+10666.7

So,

v=sqrt((12666.6)/5t+10666.7)

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

(5-0)bar v=int_0^5sqrt((12666.6)/5t+10666.7)dt

5 barv=[((12666.6)/5t+10666.7)^(3/2)/(3/2*12666.6/5)]_0^5

=(((12666.6)/5*5+10666.7)^(3/2)/(3800))-(((12666.6)/5*0+10666.7)^(3/2)/(3800))

=23333.3^(3/2)/3800-10666.7^(3/2)/3800

=648

So,

barv=648/5=129.6ms^-1

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