Question

A 4.0 kilogram object is dropped from a height of 1.0 meter onto a spring with a spring constant of 850 newtons per meter. Approximately what distance will the spring compress when the object lands on it? (1) 3.0 dm

(2) 9.0 cm

(3) 9.0 mm

(4) 2.0 dm

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Answer 1

Answer:

(1) 3.0 dm

Explanation:

Due to the law of conservation of energy, the initial gravitational potential energy of the object will be all converted into elastic potential energy of the spring as the object hits the spring:

$mgh = \frac{1}{2}kx^2$

where

m = 4.0 kg is the mass of the object

g = 9.8 m/s^2 is the acceleration due to gravity

h = 1.0 m is the initial height of the object

k = 850 N/m is the spring constant

x is the compression of the spring

Solving the equation for x, we find:

$x=\sqrt{\frac{2mgh}{k}}=\sqrt{\frac{2(4.0 kg)(9.8 m/s^2)(1.0 m)}{850 N/m}}=0.30 m=3.0 dm$

Answer 2

Option (1) is correct. The distance will the spring compress when the object lands on it will be 3.0 dm.

What is the spring constant?

Spring constant is defined as the ratio of force per unit displaced length. The force required to extend or compress a spring by some distance scales is known as the spring force.

The given data in the problem is;

m is the  mass of the object = 4.0 kg

g  is the acceleration due to gravity= 9.8 m/s²

h is the initial height of the object = 1.0 m

K is the spring constant = 850 N/m

x is the compression of the spring=?

According to the law of conservation of energy, the potential energy gets converted into the spring force. So that;

Potential energy=kinetic energy

$\rm mgh=\frac{1}{2}kx^2 \\\\ \rm x=\sqrt{\frac{2mgh}{K} } \\\\ \rm x=\sqrt{\frac{2\TIMES4.0 \TIMES 9.81\TIMES 1.0}{850} }\\\\ \rm x=0.30m\\\\ \rm x=3.0\ dm$

Hence Option (1) is correct. The distance will the spring compress when the object lands on is 3.0 dm.

TO learn more about the spring constant refer to the link;

https://brainly.com/question/4291098