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Pavlova-9 [17]
2 years ago
11

How far will 20 N of force stretch a spring with a spring constant of 140 N/m?

Physics
1 answer:
densk [106]2 years ago
7 0

Answer:

0.143 m

Explanation:

The relationship between force applied on a string and stretching of the spring is given by Hooke's law:

F=kx

where

F is the force exerted on the spring

k is the spring constant of the spring

x is the stretching of the spring from its equilibrium position

In this problem, we have:

F = 20 N is the force applied on the spring

k = 140 N/m is the spring constant

Solving for x, we find how far the spring will stretch:

x=\frac{F}{k}=\frac{20}{140}=0.143 m

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The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static fricti
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This question is incomplete, the complete question;

The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static friction at A is μ = 0.4.

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Answer:

- the magnitude of force at point A is 79.1033 lb

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Explanation:

Given that;

∑'MA = 0

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we substitute in our values

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Net force at A will be

FA' = √( NA² + FA²)

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we substitute in our values

FA' = √( (76)² + (21.9393)²)

= √( 5776 + 481.3328)

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FA' = 79.1033 lb

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FA_max = μ × NA

so, FA_max = 0.4 × 76

FA_max = 30.4 lb

So by comparing, we can easily see that the actual friction force required for keeping the the ladder stationary i.e (FA) is less than the maximum possible friction available at point A.

Therefore since FA < FA_max; Ladder WILL NOT slip

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