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A 1.89 kg object on a frictionless horizontal track is attached to the end of a horizontal spring whose force constant is 4.77 N/m. The object is displaced 5.56 m to the right from its equilibrium position and then released, which initiates simple harmonic motion. What is the magnitude of the force acting on the object 3.96 s after it is released

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Khoso123

Answer:

Magnitude F(t)=26.6 N

Direction: -x

Explanation:

Given data

Spring constant K=4.77 N/m

Mass m=1.89 kg

Displace A=5.56m

Time t=3.96s

To find

Magnitude of force F

Solution

The angular frequency is given as

[tex]w=\sqrt{\frac{K}{m} } \\w=\sqrt{\frac{4.77N/m}{1.89kg} }\\w=1.59rad/s[/tex]

Force on object is

[tex]F(t)=-mAw^{2}Cos(wt)[/tex]

Substitute given values

So

[tex]F(t)=-(1.89kg)(5.56m)(1.59rad/s)^{2}Cos(1.59*3.96)\\F(t)=-26.6N[/tex]

So

Magnitude F(t)=26.6 N

Direction: -x

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