Question

In a shipping yard, a crane operator attaches a cable to a 1240-kg shipping container and then uses the crane to lift the container vertically at a constant velocity for a distance of 37 m. Determine the amount of work done by each of the following.
(a) the tension in the cable? (J)
(b) the force of gravity? (J)

Answer 1

Answer:

(a) W = 449624 J or 449.624 kJ

(b) W = 449624 J or 449.624 kJ

Explanation:

Parameters given:

mass of container = 1240 kg

vertical distance moved = 37 m

(a) The amount of work done by Tension is given as:

W = T * d

where T = Tension in the cable;

d = distance moved by container.

The Tension in the cable is given as:

T = mg - ma

where w = weight of the container

g = acceleration due to gravity

a = normal acceleration

But a = 0, since the crane is accelerating due to gravity, being lifted vertically upwards. Hence, Tension is:

=> T = 1240 * 9.8

T = 12152 N

Hence, work done  by Tension:

W = T * d

W = 12152 * 37

W = 449624 J or 449.624 kJ

(b) The crane operator is lifting the container vertically upward. In this type of case, the Tension is equal to the weight or force of gravity.

This means that the work done by the Tension will be the same as the work done by the force of gravity.

Hence, work done by force of gravity = 449624 J or 449.624 kJ

Answer 2

Answer:

Explanation:

Given

mass of ship =1240 kg

height=37 m

(a)Work done by Tension in cable+work done by gravity =0

Work done by gravity =change in potential energy of mass$=-1240\times 9.8\times 37=-449.624 kJ$

Thus work done by tension=449.624 kJ

(b)work done by force of gravity$=-mg\times h=-1240\times 9.8\times 37$=-449.624 kJ