Question

A lift cage of mass 500 kg accelerates upwards from rest to a velocity of 6 ms−1 whilst travelling a distance of 12 m. The frictional resistance to motion is 200 N. Making use of the principle of conservation of energy, determine:

i) the work done

ii) the tension in the lifting cable

iii) the maximum power developed.

Answer 1

m = 500 kg − mass of the cage;

𝑣 = 6 m/s − final velocity of the cage;

h = 12m − distance traveled by the cage;

F = 200N − frictional resistance to motion;

t − time of travelling;

a − acceleration of the cage.

i) The lift, after accelerating h upwards gains:

- its potential energy $E_p =mgh,$

- its kinetic energy $E_k=mv^2/2,$

- in addition its motor needs to work against friction, $W_F=Fh.$

The sum of these 3 energies is equal to the driving motor's work

$$W=E_p+E_k+W_F$$

$$=mgh+mv^2/2+Fh$$

$$W=500kg(9.81m/s^2)12m+500kg(6m/s)^2/2$$

$$+200N(12m)=70260\ J=70.26\ kJ$$

ii) The whole amount of work $(W)$ is done by the tension force $\vec N$ (pointing upwards) on the distance $h.$ Because $W=N\cdot h,$ we've got:

$$N=\dfrac{70260\ J}{12m}=5855\ N$$

iii) Rate equation for the cage:

$$0=\text{v}-at$$

Equation of motion for the cage:

$$s=\dfrac{at^2}{2}$$

$$s=\dfrac{(\dfrac{\text{v}}{t})t^2}{2}$$

$$t=\dfrac{2s}{\text{v}}$$

The motor's power P equals " work / time". Therefore:

$$P=\dfrac{W}{t}=\dfrac{W\text{v}}{2s}$$

$$P=\dfrac{70260\ J(6m/s)}{2(12m)}=17565\ W=17.565\ kW$$