Asked by Technoboi11
A single force acts on a 5.0 kg particle-like object in such a way that the position of the object as a function of time is given by x = 3.0t - 4.0t 2 + 1.0t 3, with x in meters and t in seconds. Find the work done on the object by the force from t = 0 to t = 2.0 s.
find the position at at t=2 and t=0. The difference in those is the displacement.
Work= force times displacement.
So when t=0, x=0.
When t=2, x=-2
displacement= -2-0 = -2m
W=Fd
F=ma=(5)(9.8)=49
W=(49 N)(-2 m)
W= -98 J
Is this correct??
find the position at at t=2 and t=0. The difference in those is the displacement.
Work= force times displacement.
So when t=0, x=0.
When t=2, x=-2
displacement= -2-0 = -2m
W=Fd
F=ma=(5)(9.8)=49
W=(49 N)(-2 m)
W= -98 J
Is this correct??
Answers
Answered by
fuad
no>>>
find the third divetive then substute the a in F=ma< and you will get the force then find the destanse moved then
W=F x
find the third divetive then substute the a in F=ma< and you will get the force then find the destanse moved then
W=F x
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