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A particle of mass m moves along the x axis. its position varies with time according to x = (8 m / s³ )t³ + (−3 m / s² )t² . What is the work done by the force from t = 0 to t = t₁ ?

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Final answer:

To calculate the work done by the force from t = 0 to t = t₁, take the integral of the dot product of the force and the displacement function. The dot product can be split into the x and y components, and then integrated separately. The final result is 6 t₁².

Step-by-step explanation:

The work done by a force can be calculated by taking the integral of the dot product between the force and the displacement. In this case, we have the force F₁ = (3 N)Î + (4 N)Ĵ, and the displacement is given as a function of time x = (8 m / s³ )t³ + (−3 m / s² )t². To find the work done from t = 0 to t = t₁, we need to calculate the integral of the dot product of F₁ and dx from t = 0 to t = t₁. The dot product will be:

Work = ∫F₁·dx = ∫(3 N)ηdx + ∫(4 N)Ĵηdx = ∫(3 N)(8 m / s³)tdt + ∫(4 N)(−3 m / s²)tdt

Work = (3 N)(8 m / s³)∫t dt + (4 N)(−3 m / s²)∫t dt

Work = (3 N)(8 m / s³)[(1/2)t²] + (4 N)(−3 m / s²)[(1/2)t²]

Work = 12 t² - 6 t²

Work = 6 t²

To find the work done from t = 0 to t = t₁, substitute t₁ into the equation:

Work = 6 t₁²

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