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Homework Statement
A force acts on a 2.80 kg particle in such a way that the position of the particle as a function of time is given by x = (3.0 m/s)t - (4.0 m/s^2)t^2 + (1.0 m/s^3)t^3. Find the work done by the force during the first 4.0s.
Homework Equations
W = \int_{t_i}^{t_f} F(t) dt
F = ma
The Attempt at a Solution
W = \int_0^4 F(t) dt = \int_0^4 ma(t) dt = mv(4) - mv(0) = m(x'(4) - x'(0))
x'(t) = 3 - 8t + 3t^2
x'(4) = 19
x'(0) = 3
W = 2.8 * 16 = 44.8J
The book is saying it is 493J