Answer:
B) 1/5 ba^2 T^5
Explanation:
The dissipated energy is given by the work done over the object by the force F=-bv. The work is given by the following formula:
[tex]dW=Fdx[/tex]
you derivative the function f(x) and replace v by the derivative dx/dt you obtain:
[tex]v=\frac{dx}{dt}=at^2\\\\dx=at^2dt\\\\W=\int_0^{T} Fdx=-\int_0^Tvbdx=-\int_0^Tb(at^2)(at^2dt)\\\\W=-ba^2\frac{T^5}{5}=-\frac{1}{5}ba^2T^5[/tex]
hence, the dissipated energy is 1/5 ba^2 T^5
The dissipated energy by the given force during the time interval from t=0 and t=T is [tex]\bold {-\dfrac 15ba^2T^5}[/tex].
Since, the dissipated energy can be consider as work done on the object,
The work formula ,
dW = Fdx
Where,
dW - work at instance
F - force applied object
dx - displacement
Derivative the function f(x),
[tex]\bold {v = at^2} \\\\\bold{\dfrac {dx}{dt} = at^2 }\\\\\bold {dx = at^2 .dt}[/tex]
Now integrate,
[tex]\bold {W = \int\limits^t_0 {F} \, dx }[/tex]
Since, F = -bv
[tex]\bold {W = \int\limits^t_0 {-bv} \, dx }[/tex]
[tex]\bold {W = \int\limits^t_0 {-b(at^2)} \, at^2(at^2) (at^2dt)}\\\\\bold {W = -ba^2 \dfrac {T^5}{5} }\\\\\bold {W =-\dfrac 15ba^2T^5}[/tex]
Therefore, the dissipated energy by the given force during the time interval from t=0 and t=T is [tex]\bold {-\dfrac 15ba^2T^5}[/tex].
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