Question

A lift cage of mass 540 kg accelerates upwards from rest to a velocity of 6 ms−1 whilst travelling a distance of 13m. The frictional resistance to motion is 220 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

A lift cage of mass $540\,\mathrm{kg}$ accelerates upwards from rest to a velocity of 6 ms⁻¹ whilst travelling a distance of 13 m. The frictional resistance to motion is 220 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

**Solution:**

$$\begin{array}{l} m = 540\,\mathrm{kg} - \text{mass ofhte cage}; \\ v = 6\,\frac{\mathrm{m}}{\mathrm{s}} - \text{final velocity of the cage}; \\ h = 13\,\mathrm{m} - \text{distance traveled by the cage}; \\ F = 220\,\mathrm{N} - \text{frictional resistance to motion}; \\ t - \text{time of travelling}; \\ a - \text{acceleration of the cage}; \end{array}$$

**Part I**

The lift, after accelerating h upwards gains:

- its potential energy $E_{p} = mgh$

- its kinetic energy $E_{k} = \frac{mv^{2}}{2}$

- 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

$$W = E_{p} + E_{k} + W_{F} = mgh + \frac{mv^{2}}{2} + Fh$$

Let's calculate W (assuming $g = 9.8\,\frac{\mathrm{m}}{\mathrm{s}^2}$)

$$W = 540\,\mathrm{kg} \cdot 9.8\,\frac{\mathrm{m}}{\mathrm{s}^2} \cdot 13\,\mathrm{m} + \frac{540\,\mathrm{kg} \cdot \left(6\,\frac{\mathrm{m}}{\mathrm{s}}\right)^2}{2} + 220\,\mathrm{N} \cdot 13\,\mathrm{m} = 81.4\,\mathrm{kJ}$$

**Part 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:

$$N = \frac{W}{h} = \frac{81.4\,\mathrm{kJ}}{13\,\mathrm{m}} = 6262\,\mathrm{N}$$

**Part III**

Rate equation for the cage:

$$0 = V - a t$$ $$t = \frac{V}{a}$$

Equation of motion for the cage:

$$S = \frac{a t^{2}}{2}$$

(1) in (2):

$$\begin{array}{l} 2S = a \cdot \left(\frac{V}{a}\right)^2 \\ 2S = \frac{V^2}{a} \\ a = \frac{V^2}{2S} = \frac{\left(6 \frac{\text{m}}{\text{s}}\right)^2}{2 \cdot 13 \text{m}} = 1.4 \frac{\text{m}}{\text{s}^2} \Rightarrow \\ t = \frac{V}{a} = \frac{6 \frac{\text{m}}{\text{s}}}{1.4 \frac{\text{m}}{\text{s}^2}} = 4.3 \text{s} \end{array}$$

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

$$P = \frac{W}{t} = \frac{81.4 \text{ kJ}}{4.3 \text{s}} = 18.9 \text{kW}$$

Answer: I) 81.4 kJ II) 6262N III) 18.9kW.