When a 0.50 kg-object is attached to a vertically supported spring, it stretches 0.10 m from the equilibrium position. Find k for the spring.
k = M*g/(length change) = 4.9/0.1
= 490 N/m
To find the spring constant (k) for the spring, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position.
Hooke's Law formula is given as:
F = -kx
Where:
F = Force exerted by the spring
k = Spring constant
x = Displacement
From the problem, we know that the object has a mass of 0.50 kg and it stretches the spring by 0.10 m. We need to find the value of k.
The force exerted by the spring (F) can be calculated using the formula: F = m * g
Where:
m = Mass of the object
g = Acceleration due to gravity (approximately 9.8 m/s²)
Substituting the given values, we have:
F = 0.50 kg * 9.8 m/s²
Next, we equate the force exerted by the spring to the negative of the displacement multiplied by the spring constant:
F = -kx
0.50 kg * 9.8 m/s² = -k * 0.10 m
Simplifying the equation, we get:
4.9 N = -k * 0.10 m
To find k, we can rearrange the equation as follows:
k = (4.9 N) / (0.10 m)
Evaluating the expression, we find:
k = 49 N/m
Therefore, the spring constant (k) for the spring is 49 N/m.