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A particle moving in the +ve x direction has an acceleration, $a = 100 - 4v^2 m/s^2.$ Determine the time interval and displacement of a particle when speed changes from 1 m/s to 3 m/s.
2 Answers
written 7.3 years ago by |
FBD of the lift
Using kinematic equation
$ v^2=u^2+2as$
Given that $v=3m/s, u=0, s=4m,$ we have,
$ 3^2=0+(2×4)a \\ a = \dfrac98 m/s^2 $
By Newton’s Law,
$ma=T-mg \\ \therefore T=m(a+g)=750(\dfrac 98+9.81) \\ T=8201.25N \space or\space T=8.201kN -----Ans$
written 7.3 years ago by |
FBD of the lift
Using kinematic equation
$ v^2=u^2+2as$
Given that $v=3m/s, u=0, s=4m,$ we have,
$ 3^2=0+(2×4)a \\ a = \dfrac98 m/s^2 $
By Newton’s Law,
$ma=T-mg \\ \therefore T=m(a+g)=750(\dfrac 98+9.81) \\ T=8201.25N \space or\space T=8.201kN -----Ans$