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Difficulty: MediumMechanics and Motion

A mechanical crane lifts a cargo container with a mass of 500 kg500\text{ kg} vertically upward. If the crane's cable exerts a constant upward tension of 6400 N6{}400\text{ N} and the acceleration due to gravity is 9.8 m/s29.8\text{ m/s}^2, what is the upward acceleration of the container?

  1. A
    12.8 m/s212.8\text{ m/s}^2
  2. 3 m/s23\text{ m/s}^2Answer
  3. C
    11.8 m/s211.8\text{ m/s}^2
  4. D
    0.3 m/s20.3\text{ m/s}^2

Answer

3 m/s23\text{ m/s}^2
The container experiences two vertical forces: an upward tension of 6400 N6{}400\text{ N} and a downward gravitational force (weight) of W=mg=500 kg×9.8 m/s2=4900 NW = mg = 500\text{ kg} \times 9.8\text{ m/s}^2 = 4{}900\text{ N}. The net upward force is 6400 N−4900 N=1500 N6{}400\text{ N} - 4{}900\text{ N} = 1{}500\text{ N}. Applying Newton's second law (a=Fnet/ma = F_{\text{net}} / m) gives an acceleration of 1500 N/500 kg=3 m/s21{}500\text{ N} / 500\text{ kg} = 3\text{ m/s}^2.

Step-by-Step Solution

1
Calculate the downward gravitational force (weight) acting on the container.
W=mg=500 kg×9.8 m/s2=4900 NW = mg = 500\text{ kg} \times 9.8\text{ m/s}^2 = 4{}900\text{ N}
Weight is the force exerted on the mass by gravity.
2
Find the net upward force by subtracting weight from the upward cable tension.
Fnet=Ftension−W=6400 N−4900 N=1500 NF_{\text{net}} = F_{\text{tension}} - W = 6{}400\text{ N} - 4{}900\text{ N} = 1{}500\text{ N}
The acceleration depends on the resultant unbalanced force acting on the object.
3
Apply Newton's second law (a=Fnet/ma = F_{\text{net}} / m) to determine the acceleration.
a=1500 N500 kg=3 m/s2a = \frac{1{}500\text{ N}}{500\text{ kg}} = 3\text{ m/s}^2
Acceleration equals net force divided by total mass.

Key Concept

Newton's Second Law and Net Vertical Force
Estimated Time:1m 0s
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Topics

Mechanics and Motion

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