- #1
SonRuy
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An applied force varies with position according to F = k1 xn− k2, where n = 3,
k1 = 2.4 N/m3, and k2 = 56 N.
How much work is done by this force on an object that moves from xi = 5.13 m to xf = 28.2 m?
Answer in units of kJ
i keep using the integration equation: w = ∫f(x) dx
where k1x^4/4 - 56x is the new equation which i plug 28.2m and 5.13m and subtract
i get 377736.527
but its wrong, help please?
k1 = 2.4 N/m3, and k2 = 56 N.
How much work is done by this force on an object that moves from xi = 5.13 m to xf = 28.2 m?
Answer in units of kJ
i keep using the integration equation: w = ∫f(x) dx
where k1x^4/4 - 56x is the new equation which i plug 28.2m and 5.13m and subtract
i get 377736.527
but its wrong, help please?