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olga nikolaevna [1]
3 years ago
9

An object of mass 0.40 kg, hanging from a spring with a spring constant of 8.0 N/m, is set into an up-and-down simple harmonic m

otion. What is the magnitude of the acceleration of the object when it is at its maximum displacement of 0.10 m?
Physics
1 answer:
Sergeeva-Olga [200]3 years ago
5 0

Answer:

a = 2 m/s2

Explanation:

we know from newtons 2nd law

F = ma.

we also know that from hookes law we have

F = kx

equate both value of force to get value of acceleration

kx = ma,

where,

k is spring constant = 8.0 N/m

x is maximum displacement  0.10 m

m is mass of object 0.40 kg

a = \frac{kx}{m}

     = \frac{8 *0 .10}{0.40}

a = 2 m/s2

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Answer:

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T₂ : Wire tension forming angle of 40° with horizontal

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We replace T₁ of the equation (1) in the equation (2)

1.244T₂*sin52° + T₂*sin40° - 98 = 0

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1.62T₂ = 98

T₂ = 98 / 1.62

T₂ = 60.49 N

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T₁= 1.244*60.49 N

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