A 5.60-kg particle
moves along the x axis.
Its position varies with time according to x = t
+ 4.0t3, where x
is in meters and t
is in seconds.
a | = | |
F | = |
x = t + 4.0t^3
V=differentiation(x)=1+12*t^2
a=24t
a)Kinetic energy=0.5*m*v^2
=0.5*5.6*(1+12*t^2)^2
=2.8*(1+12*t^2)^2
b)a=24t
F=Ma=5.6*24t=134.4t
c)power=FV=(134.4t)(1+12*t^2)
134.4t+1612.6*t^3
d)work done=
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