Question

An object moves in one dimension according to the function x(t)=1/3at³, where a is a positive constant with units of ms³. During this motion, a force F =−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− 1/9ba²T⁵
B− 1/5ba²T⁵
C− 1/3ba²T⁵
D− ba²T⁵

Answer 1

To calculate the energy dissipated by a force in the given scenario, the velocity function is derived from the position function, and then the work-energy principle is applied, resulting in the integral of the force with respect to displacement. Option B is correct answer.

Step-by-step explanation:

The student is asking about the energy dissipated by a force given a specific one-dimensional motion of an object. To find the energy dissipated by the force F = -bv over a time interval from t=0 to t=T, where v is the velocity of the object, we start by finding the velocity as a function of time using the given position function x(t) = (1/3)at^3. The velocity v is the derivative of x(t), so v(t) = dx(t)/dt = at^2.

Then we calculate the energy dissipated using the work-energy principle, with work done by the force F being equal to the change in kinetic energy, which is in this case lost to the force F.

The work done by this force over the time interval from t=0 to t=T is given by the integral of F with respect to x, which translates to the integral of F times v with respect to t. Substituting the expressions for F and v, we get W = ∫ F dx = ∫ (-bv) (v dt) = -b ∫ (at^2)^2 dt = -b ∫ a^2t^4 dt. Evaluating this integral from t=0 to t=T yields W = - (1/5)ba^2T^5, which is option B.