An object has a mass of 4 kg. The object's kinetic energy uniformly changes from 104 KJ to 72KJ over t in [0,5s]. What is the average speed of the object?

1 Answer
Narad T.
Mar 15, 2017

The average speed is =209.5ms^-1

Explanation:

The kinetic energy is

KE=1/2mv^2

The initial velocity is =u_1

1/2m u_1^2=104000J

The final velocity is =u_2

1/2m u_2^2=72000J

Therefore,

u_1^2=2/4*104000=52000m^2s^-2

and,

u_2^2=2/4*72000=36000m^2s^-2

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

The points are (0,52000) and (5,36000)

The equation of the line is

v^2-52000=(36000-52000)/5t

v^2=-3200t+52000

So,

v=sqrt((-3200t+52000)

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

(5-0)bar v=int_0^5sqrt(-3200t+52000) dt

5 barv=[(-3200t+52000)^(3/2)/(3/2*-3200)]_0^5

=((-3200*5+52000)^(3/2)/(-48000))-((-3200*0+52000)^(3/2)/(-48000))

=36000^(3/2)/-4800-52000^(3/2)/-4800

=1/4800(52000^(3/2)-36000^(3/2))

=1047.4

So,

barv=1047.4/5=209.5ms^-1

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