草稿纸
全日制-M3-第4课-出门测-2023
题量：4题
时间
45分钟
满分：20分
〇゚剩余
1.(5.0分) The length of an elastic spring is reduced to 0.8 m when a force of 20 N compresses it. Given that the modulus of elasticity of the spring is 25 N , find its natural length.
A 0
B 2

C 4

D 6

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Question

草稿纸
全日制-M3-第4课-出门测-2023
题量：4题
时间
45分钟
满分：20分
〇゚剩余
1.(5.0分) The length of an elastic spring is reduced to 0.8 m when a force of 20 N compresses it. Given that the modulus of elasticity of the spring is 25 N , find its natural length.
A 0
B 2

C 4

D 6

不会做? 点这里?

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ We start with the given mass of the particle $ m = 0.25 \, \text{kg} $ and the relationship $ F = ma $. We need to find the acceleration $ a $ when $ x = 5 $. ‖ ‖ step 2 ⋮ We know that $ \frac{dv}{dx} = 3 $. To find $ a $, we use the chain rule: $ a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = \frac{dv}{dx} \cdot v $. ‖ ‖ step 3 ⋮ At $ x = 2 $, we have $ v = 3 $. Thus, when $ x = 5 $, we can find $ v $ using the relationship $ v = 3 + 3(x - 2) = 3 + 3(5 - 2) = 3 + 9 = 12 $. ‖ ‖ step 4 ⋮ Now substituting $ v = 12 $ into the acceleration formula: $ a = 3 \cdot 12 = 36 $. ‖ ‖ step 5 ⋮ Finally, we calculate the force: $ F = ma = 0.25 \cdot 36 = 9 \, \text{N} $. ‖ ***A*** ∻Key Concept∻ ⚹Newton's Second Law⚹ ∻Explanation∻ ⚹The force acting on an object is equal to the mass of the object multiplied by its acceleration.⚹◊What is the formula to calculate the potential energy stored in a compressed spring using its natural length?⍭ Generate me a similar question◊

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