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PilotLPTM [1.2K]
3 years ago
14

What happens when you push a spring? How is this different than pulling it? (Hooke’s Law)

Physics
2 answers:
Yuki888 [10]3 years ago
7 0

A spring is an object that can be deformed by a force and then return to its original shape after the force is removed.Springs come in a huge variety of different forms, but the simple metal coil spring is probably the most familiar. Springs are an essential part of almost all moderately complex mechanical devices; from ball-point pens to racing car engines.

Furkat [3]3 years ago
4 0

When you push a spring you are putting force on it, or rather forcing the spring onto its self. This is different than pulling the spring because when you push it and let go it will spring but if you pull it its would not move as far as it would if you push it. Pulling the spring would most likely deform the spring will pushing it would cause it to spring back into place.

<u>Hooke's law states that the force that extends or compresses a spring by some distance scales linearly with respect to that distance.</u>

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Answer:

The answer is "151.25 J and -547.64 J".

Explanation:

u = 0\\\\v = 2.35\  \frac{m}{sec}\\\\d = 5.0 \ m\\\\

Using formula:

v^2 = u^2 + 2 \times a \times d\\\\2.35^2 = 0^2 + 2 \times a \times 5\\\\a = \frac{2.35^2}{10} \\\\

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F_{net} = m \times a\\\\F_{net} = 55 \times 0.55 = 30.25\ N\\\\

Calculating the Work by net force

W = F_{net}\times d\\\\W = 30.25 \times 5 = 151.25 \ J\\\\

The above work is converted into thermal energy.

Now,

F_{net} = F_p - F_f\\\\F_p = 140 \ N\\\\F_f = u_k\times m \times g = u_k \times 55 \times 9.81\\\\F_f = 539.55 \times u_k\\\\30.25 = 140 - u_k \times 55 \times 9.81\\\\u_k = \frac{(140 - 30.25)}{(55\times 9.81)}\\\\uk = 0.203 = \text{Coefficient of friction}\\\\W_f = -F_f \times d\\\\W_f = -0.203 \times 55 \times 9.81 \times 5\\\\Work\ done\  by\ friction = -547.64 \ J

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