I 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, the answer I get is .5m which doesn't seem right.
Thank you
The word equilibrium is confusing here, so I am assuming as you did that the equilibrium length for a horizontal spring is its stretched length.
You used the correct relationship: max PE equals starting KE.
then if follows
x= v/sqrt k
I don't get your answer, I get less.
did u get .05m? i don't why i am getting that now
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Yes, you should get 0.05 m. The equation is 1/2mv^2 = 1/2kx^2, so x = v/sqrt(k). Plugging in the values, you get x = 5/sqrt(200) = 0.05 m.
To find the maximum elongation of the spring, you can use the equation you mentioned, which is the conservation of mechanical energy equation:
1/2mv^2 = 1/2kx^2
where m is the mass of the block (2 kg), v is the velocity of the block (5 m/s), k is the spring constant (200 N/m), and x is the maximum elongation of the spring.
To solve for x, rearrange the equation:
x^2 = (mv^2) / k
x^2 = (2 kg * (5 m/s)^2) / 200 N/m
x^2 = (2 kg * 25 m^2/s^2) / 200 N/m
x^2 = 0.25 m^2
Taking the square root of both sides gives:
x = sqrt(0.25) m
x = 0.5 m
Therefore, the maximum elongation of the spring is 0.5 meters.
It seems that your initial calculation of 0.5 m is correct. Make sure to double-check your calculations to ensure accuracy.
To find the maximum elongation of the spring, we can use the equation:
Potential Energy (PE) = Kinetic Energy (KE)
Given:
Mass (m) = 2 kg
Spring constant (k) = 200 N/m
Speed (v) = 5 m/s
The equation for potential energy in a spring is:
PE = 1/2 * k * x^2
The equation for kinetic energy is:
KE = 1/2 * m * v^2
Setting PE equal to KE:
1/2 * k * x^2 = 1/2 * m * v^2
Substituting the given values:
1/2 * 200 * x^2 = 1/2 * 2 * 5^2
Simplifying:
100 * x^2 = 10 * 25
100 * x^2 = 250
Dividing both sides by 100:
x^2 = 2.5
Taking the square root of both sides:
x = √2.5
Simplifying:
x ≈ 1.58 m
Therefore, the maximum elongation of the spring is approximately 1.58 m.