Question

An object moves in one dimension according to the function x(t)=13at3, where a is a positive constant with units of ms3. During this motion, a force F⃗ =−bv is exerted on the object, where v is the object’s velocity and b is a constant with units of kgs. Which of the following expressions will yield the amount of energy dissipated by this force during the time interval from t=0 and t=T ?
A− 19ba2T5
B− 15ba2T5
C− 13ba2T5
D− ba2T5

Answer 1

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:

$dW=Fdx$

you derivative the function f(x) and replace v by the derivative dx/dt you obtain:

$v=\frac{dx}{dt}=at^2\\\\dx=at^2dt\\\\W=\int_0^{T} Fdx=-\int_0^Tvbdx=-\int_0^Tb(at^2)(at^2dt)\\\\W=-ba^2\frac{T^5}{5}=-\frac{1}{5}ba^2T^5$

hence, the dissipated energy is 1/5 ba^2 T^5