A spring is hung from a ceiling, and an object attached to its lower end stretches the spring by a distance of 5.00 cm from its unstretched position when the system is in equilibrium. If the spring constant is 54.1 N/m, determine the mass of the object.
ur bad
force= k x^2
mg=kx^2 solve for m.
To determine the mass of the object, we need to use Hooke's law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position.
Hooke's Law: F = -kx
Where:
F is the force exerted by the spring,
k is the spring constant, and
x is the displacement from the equilibrium position.
In this case, the displacement (x) is given as 5.00 cm = 0.050 m, and the spring constant (k) is given as 54.1 N/m.
Using the formula F = -kx, we can rearrange it to solve for the force (F) exerted by the spring:
F = -kx
F = -(54.1 N/m)(0.050 m)
F = -2.70 N
The negative sign indicates that the force is in the opposite direction of the displacement.
To determine the mass of the object, we can use Newton's second law, which states that the force acting on an object is equal to its mass times acceleration.
Newton's Second Law: F = ma
Rearranging the formula to solve for mass (m):
m = F / a
In this case, the force (F) is given as -2.70 N. Since the system is in equilibrium, the acceleration (a) is zero.
So, m = -2.70 N / 0 (which is 0 because the acceleration is zero)
m = 0 kg
Therefore, the mass of the object is 0 kg.