have a question involving the spring costant:
A 2-kg block is attached to a horizontal ideal spring with a spring constant of 200N/m. When the spring has its equilibrium length the block is given a speed of 5 m/s. What is the maximum elongation of the spring? The answer is supposed to be in m.
I used 1/2mv^2=1/2KX^2
When I solved I got .25m, this time. This is not even close to what the book gives.
x^2= mv^2/K
I made an error before, forgot m.
x= v sqrt (m/k)
I get .5m
How does the book get an answer of .05m? I'm confused
x=v sqrt (m/K) = 5* sqrt (2/200)
=5 sqrt (1/100)= .5m
Could it be the book is wrong? Heaven forbid. Trust the force, as Obiwan Kenobi stated.
To find the maximum elongation (x) of the spring, you correctly used the equation x = v * sqrt(m/K), where v is the initial velocity of the block, m is the mass of the block, and K is the spring constant.
In this case, the mass of the block is 2 kg and the spring constant is 200 N/m. So, plugging in these values, we have:
x = 5 * sqrt(2/200)
x = 5 * sqrt(1/100)
x = 5 * (1/10)
x = 0.5 m
Therefore, the correct answer to the maximum elongation of the spring is indeed 0.5 m, not 0.05 m. It seems there may be an error in the book.
Remember to always double-check your calculations and formulas, as well as make sure you have input the correct values. In this case, it appears that your calculation is correct, and the book might have a mistake. Trusting and verifying your work is essential to ensure accuracy.
Based on the given values, it looks like you have done the calculations correctly. The maximum elongation of the spring can be calculated using the formula x = v * sqrt(m/k), where x is the maximum elongation, v is the velocity, m is the mass, and k is the spring constant.
Substituting the values, we get:
x = 5 * sqrt(2/200) = 5 * sqrt(0.01) = 5 * 0.1 = 0.5m
From your calculations, it seems that you also obtained the same answer of 0.5m. Therefore, it appears that there might be an error in the book, and the correct answer should be 0.5m, not 0.05m.