# A particle of mass m moves along the x axis. Its position varies with time according to x=(6m/s^3) t^3 + (-6m/s^2) t^2. What is the work done by the force from t=0 to t=t1?

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### 1 Answer

A particle of mass m moves along the x-axis such that x = 6m*t^3 - 6m*t^2

The velocity of the particle is x' = 18m*t^2 - 12mt

The acceleration of the particle is x'' = 36mt - 12m

The force that moves the particle is m*x'' = 36m^2*t - 12m^2

Work done by a force exerted over a distance d is the product of the force and the displacement.

Here, the work done by the force from time t = 0 to t = t1 is the definite integral of 36m^2*t - 12m^2 dt between 0 and t1

W = 18m^2*(t1^2 - 0^2) - 12m^2*(t1 - 0)

=> 18m^2*t1^2 - 12m^2*t1

**The work done by the force from t = 0 to t = t1 is 18m^2*t1^2 - 12m^2*t1**

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