Question

A 2000-kg elevator accelerates upward uniformly at 1 m/s² from a stopped position. (a) What is the tension in the lifting cable in kN? (b) At the end of 4 s of operation, what will be the kinetic energy and the change of potential energy in kJ? Local gravity acceleration is 9.70 m/s²

T =

KE = PE =

Answer 1

Solution;

(a)Tension in lifting cable;

T=mg+ma

T is tension force;

m is mass;

g gravitational acceleration;

a is elevator acceleration.

T=(2000×9.70)+(2000×1)

T=21400N

T=21.4kN

(b)Kinetic energy and potential energy;

Given;

t=4s

u=0(initial velocity)

Find the displacement(s) after 4s;

$s=ut+\frac12at^2$

$s=(0×4)+(\frac12×1×4^2)$

s=8m

Kinetic energy KE;

$KE=\frac12mv^2$

$v=at=1×4=4m/s$

$KE=\frac12×2000×4^2$

$KE=16kJ$

Change of potential energy;

$\Delta PE=mg\Delta h$

$\Delta PE=2000×9.7×(8-0)$

$\Delta PE=155.2kJ$