Question

A vertical spring with k=490N/m is standing on the ground. You are holding a 5.0kg block just above the spring, not quite touching it. Part A How far does the spring compress if you let go of the block suddenly? Express your answer to two significant figures and include the appropriate units.

Answer 1

Answer:

Compression in the spring, x = 0.20 m

Explanation:

Given that,

Spring constant of the spring, k = 490 N/m

Mass of the block, m = 5 kg

To find,

Compression in the spring.

Solution,

Since the block is suddenly dropped on the spring gravitational potential energy of block converts into elastic potential energy of spring. Its expression is given by :

$mgx=\dfrac{1}{2}kx^2$

Where

x is the compression in the spring

$x=\dfrac{2mg}{k}$

$x=\dfrac{2\times 5\times 9.8}{490}$

x = 0.20 m

So, the compression in the spring due to block is 0.20 meters.

Answer 2

The vertical spring with spring constant of 490N/m will compress at a distance of 0.2m.

How to calculate distance of spring?

According to this question, the following information were given:

• Spring constant of the spring, k = 490 N/m
• Mass of the block, m = 5 kg

To calculate the compression in the spring, we deduce that the gravitational potential energy of the spring converts into elastic potential energy. Its expression is given by:

mgx = ½kx²

Where;

• x = compression in the spring

x = 2mg/k

x = 2 × 5 × 9.8/490

x = 98/490

x = 0.2m

Therefore, the spring will compress at a distance of 0.2m.

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