A particle moving in the +ve x direction has an acceleration, $a = 100

1.5kviews

A particle moving in the +ve x direction has an acceleration, $a = 100 - 4v^2 m/s^2. $

written 9.4 years ago by

teamques10 ★ 70k

• modified 5.6 years ago

Determine the time interval and displacement of a particle when speed changes from 1 m/s to 3 m/s.

engineering mechanics

ADD COMMENT EDIT

1 Answer

2views

written 9.4 years ago by

teamques10 ★ 70k

FBD of the lift

enter image description here

Using kinematic equation

$$v^2=u^2+2as$$

Given that $v=3m/s, u=0, s=4m, $ we have,

$$3^2=0+(2×4)a$$

$\therefore a=\dfrac 98 m/s^2$

By Newton’s Law,

$ma=T-mg \\ \therefore T=m(a+g)=750(\dfrac 98+9.81) \\ T=8201.25N \space \space or \space \space T=8.201kN --Ans$

ADD COMMENT EDIT

Please log in to add an answer.