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A horizontal spring with a block attached to the end has a natural length of 10 cm. When the spring is attached to the ceiling, the natural length increases by 2 cm. If the spring constant of the block is 50 N/m, what is the mass of the block?(A) 0.01 kg(B) 0.05 kg(C) 0.1 kg(D) 0.5 kg

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Chapter 13

Practice Test 3

Section 1

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Hi. In the given problem, natural land of the spring Is given to be L. is equal to 10 cm and change in the land and the block is suspended with it Is given as Delta L. is equal to two cm. The spring factor of this spring is given as K. is equal to 50 Newton per meter. So Using the expression for historian Force F equals two K into delta L. And here we are using only the magnitude because otherwise it should have a negative sign also means ethical to minus K into delta L. And this restoring force F will be actually equal to the deforming force, which is nothing but the weight of bloc suspended with the spring. So M G is equal to get delta L. Will give us an expression for the mass of the block, which is scared into delta L by G. So plugging in unknown values for K. This is 50 newton per meter, multiplied by two cm. Or we can say this is doing to 10 H 1 -2 m, divided by G, which is 10 newton parameters square. So canceling this, we get the mass of the bloc suspended with a spring to be 0.1 kilogram. Hence, you can see. Yeah, corruption. See is correct. Thank you.

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