A 2.25-kg object is attached to a horizontal an ideal massless spring on a frictionless table. What should be the spring constant of this spring so that the maximum acceleration of the object will be g when it oscillates with amplitude of 4.50 cm?

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Q: When 1/5 Kg (200g) mass is applied vertically to an elastic spring, the spring is stretched 10 cm…

A: mass = 200 g x = 10 cm therefore, amplitude = 10 cm  = 0.1 musing , hooke's law F = k x mg = kx 0.2…

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A: Given, mass of block, m = 0.36 kg x = 0.028 m

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A: Since you have posted a question with multiple sub parts, we willprovide the solution only to the…

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A: In a spring mass system, spring force depends on the displacement of the mass, greater is the…

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A: The potential associated with the spring is elastic potential energy that is given as  Ep=1/2 kx2

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Q: Ax m k

A: When the spring is stretched by an amount x then the restoring force acting on it is F=kx. From the…

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A: The spring constant (denoted by k) is given here as 40 N/m. Let x denotes the compression of the…

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A: Given Data: The magnitude of mass is: m=0.92 kg The spring constant is: k=350 N/m The displacement…

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A: Given: mass of block m = 0.44 kg stretched distance x = 0.8 m force constant of spring k=4 N/m

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Q: 1. A block of mass M is pressed up against a spring of spring constant k. The block is not attached…

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Q: Part (b) Does increasing the mass at the end increase the maximum speed? O Yes O No

A: Write the expression of maximum speed. vmax=2gL1-cosθ

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A: Given data: A spring-mass system  Spring constant (k) = 39 N/m Amplitude (A) = 7 cm = 0.07 m…