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A 5000 kg elephant steps onto a larg spring and compresses it from 1 m long to 50cm long what is the spring constant of the spring

User Lovespeed
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1 Answer

5 votes

Answer: 98000 N/m.

Step-by-step explanation:

To determine the spring constant of the spring, we can use Hooke's Law, which states that the force exerted by a spring is proportional to its displacement.

Hooke's Law can be written as:

F = -k * x

Where:

F is the force applied to the spring,

k is the spring constant, and

x is the displacement of the spring from its equilibrium position.

In this case, we can assume that the displacement of the spring is the change in length from 1 m to 50 cm, which is 0.5 m (or 0.5 m - 1 m = -0.5 m if we consider compression as a negative displacement).

We can also assume that the force exerted by the spring is the weight of the elephant, which is given as 5000 kg.

Plugging these values into Hooke's Law:

5000 kg * 9.8 m/s^2 = -k * 0.5 m

Simplifying the equation:

49000 N = -k * 0.5 m

Dividing both sides by -0.5 m:

k = -49000 N / -0.5 m

k = 98000 N/m

Therefore, the spring constant of the spring is 98000 N/m.

User JTew
by
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