A 1.50 kg object hangs motionless from a spring with a force constant of k = 250 N/m. How far is the spring stretched from its equilibrium length?
Pls help
To find the distance the spring is stretched from its equilibrium length, we can use Hooke's law. Hooke's law states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position.
Mathematically, we can express Hooke's law as:
F = -kx
Where:
F is the force exerted by the spring (in newtons),
k is the force constant of the spring (in newton per meter), and
x is the displacement from the equilibrium position (in meters).
In this case, the object is hanging motionless, which means that the gravitational force acting on it is equal to the force exerted by the spring. We can set up an equation to represent this:
mg = kx
Where:
m is the mass of the object (in kilograms),
g is the acceleration due to gravity (approximately 9.8 m/s²),
k is the force constant of the spring (250 N/m), and
x is the displacement from the equilibrium position (what we want to find).
Rearranging the equation, we can solve for x:
x = mg / k
Plugging in the values given:
m = 1.50 kg
g = 9.8 m/s²
k = 250 N/m
x = (1.50 kg * 9.8 m/s²) / 250 N/m
x = 0.0588 m
Therefore, the spring is stretched approximately 0.0588 meters (or 5.88 cm) from its equilibrium length.