Question

a dock side container crane can lift a fully loaded container of mas 37,000 kg from rest to a maximum velocity of 1.2ms-1 in a distance of 8m. if the friction ristance of the crane winch gear is 2kn anbd asuming the accleration is inform determain using the princible of convertation of energy:
i work done
ii the tension in the lifting cable
the maximum power developed

Answer 1

Answer on Question #68497 – Physics – Mechanics | Relativity

A dock side container crane can lift a fully loaded container of mas $37000\,kg$ from rest to a maximum velocity of $1.2\,ms^{-1}$ in a distance of $8\,m$. if the friction distance of the crane winch gear is $2\,kN$ and assuming the acceleration is inform determinin using the principle of conversation of energy:

i. work done

ii. the tension in the lifting cable

iii. the maximum power developed

Solution.

We find the acceleration of the container:

$$a = \frac{v^2 - v_0^2}{2s} = \frac{1.2^2 - 0}{2*8} = \frac{1.44}{16} = 0.09\,ms^{-2};$$

The Newton's second law for the motion of a container (in projection on the vertical y-axis):

$\mathrm{m} * \mathrm{a} = F_1 - mg; F_1$ is the tension in the lifting cable

$$F_1 = m(a + g) = 37000 * (9.81 + 0.09) = 366300\,N \approx 366\,kN;$$

We find the maximum power developed using the principle of conservation of energy.

$$P = F * v = (F_1 + F_2) * v = (366300 + 2000) * 1.2 = 441960\,W \approx 442\,kW; F_2 \text{ is the friction distance of the crane}.$$

Answer:

ii. $F_1 \approx 366\,kN$;

iii. $P \approx 442\,kW$

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