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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 ?

User Vernonk
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Final answer:

The amount of energy dissipated by the force from t=0 to t=T is calculated using the work-energy principle. By integrating the product of the force and velocity over time, the dissipated energy is found to be 304.2a^2bT^5 joules.

Step-by-step explanation:

To find the amount of energy dissipated by the force during the time interval from t=0 to t=T, we can use the work-energy principle. The work done by the force on the object is equal to the change in kinetic energy of the object.

First, we need to find the object's velocity as a function of time (v(t)). The given position function is x(t) = 13at3. Differentiating this with respect to time, we get:

v(t) = dx/dt = d(13at3)/dt = 39at2

The force exerted on the object is F = -bv, where v is the velocity and b is a constant. Substituting v(t) into the expression for force, we have:

F(t) = -b(39at2) = -39abt2

The work done by this force over the time interval from 0 to T is the integral of the force with respect to displacement. Since we have the force as a function of time, we need to multiply it by the velocity to get the integrand in terms of displacement:

W = ∫0T F(t) · v(t) dt = - ∫0T (39abt2) (39at2) dt

W = - ∫0T 1521a2b t4 dt

Performing the integration, we find:

W = - ∫0T 1521a2b t4 dt = - [304.2a2b t5]0T

W = -304.2a2b T5

This negative value indicates that energy is being taken away from the system, i.e., dissipated due to the force. Therefore, the amount of energy dissipated is 304.2a2b T5 joules.

User Jennykwan
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