# An applied force varies with position according to F=k1 x^n-k2, where n=3, k1=3.4N/m^3, and k2=53N. How much work is done by this force on an object that moves from xi=4.73m to xf= 13m?

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### 1 Answer

An applied force varies with position according to F = k1*x^n - k2, where n = 3, k1 = 3.4 N/m^3, and k2 = 53N.

Substituting the values of k1, k2 and n gives F = 3.4*x^3 - 53

The work done by a force F is the product of the force and the displacement due to the force.

Here, as the force is not constant, it is the definite integral of (3.4*x^3 - 53) dt from x = 4.73 to x = 13

This is equal to:

(3.4/4)*(13^4 - 4.73^4) - 53(13 - 4.73)

=> 23413.07 J

=> 23.413 kJ

**The work done by the force is equal to 23.413 kJ**

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