We have thathat is the net horizontal force on the mass at t=0s and t=1s is
F=0N
F=0N respectively
From the question we are told
The position of a 3.0 kg mass is given by \(x=(2t^3−3t^2)\) m, where t is in seconds.
What is the net horizontal force on the mass at t=0s?
What is the net horizontal force on the mass at t=1s?
Generally the equation for the position of the particle is is mathematically given as
\(x=(2t^3−3t^2)\)
Therefore
\(dx/dt=v\\\\dx/dt=6t^2-6t\\\\v=6t^2-6t\)
Where
\(dv/dt=6t-6\\\\a=6t-6\)
Where
\(F=ma\\\\at t=0\\\\F=6(0)-6*3\\\\F=0N\)
\(at t=1\\\\F=6(1)-6*3\\\\F=18-18\\\\F=0\)
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