A block of ice with mass 5.80 kg is initially at rest on a frictionless, horizontal surface. A worker then applies a horizontal force \vec{F} to it. As a result, the block moves along the x-axis such that its position as a function of time is given by x(t)= ( 0.204 m/s^2) t^{2}+ ( 1.98×10−2 m/s^3) t^{3}.
Known Values from previous problems:
velocity of the object at time t = 3.80 s.
2.41 m/s
magnitude of \vec{F} at time t = 3.80 s.
4.98 N
would appreciate answer and/or explanation.
2007-03-14
14:19:32
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2 answers
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asked by
eric
2
in
Science & Mathematics
➔ Physics