A 0.150 kg particle moves along an x axis according to x(t) = −13.00 + 2.00t...

A 0.150 kg particle moves along an x axis according to x(t) = −13.00 + 2.00t + 3.50t2 − 2.50t3, with x in meters and t in seconds. In unit-vector notation, what is the net force acting on the particle at t = 3.55 s?

F with arrow = ______ N

Expert Solution

mass of the particle m = 0.15 kg

time t = 3.55 s

position vector x(t) = -13.00 + 2.00t + 3.50t² - 2.50t³,

velocity v = dx/dt

= 0 + 2.00 + 2*3.5 t - 2.5*3 t²

= 2.00 + 7 t - 7.5 t²

acceleration a = dv/dt

= 7- 15 t

force along the x-axis is

F = (ma) i^

= [(0.15 kg)( 7 - 15 t)] i^

= [(0.15 kg)( 7 - 15*3.55 s)] i^

= [-6.93 N] i^

genius_generous answered 2 years ago

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