A spring that has a force constant of 1200 N/m is mounted vertically on the ground. A block of mass 1.60 kg is dropped from rest from height of 1.50 m above the free end of the spring. By what distance does the spring compress?
the compression of the spring = change PE of the block
mg(1.5+x)=1/2 k x^2
solve for x
To find the distance by which the spring compresses, we need to consider the conservation of mechanical energy. Initially, the block has gravitational potential energy due to its height above the free end of the spring. As it falls, this potential energy is converted into the potential energy stored in the compressed spring.
To calculate the compression distance, we can use the equation for potential energy stored in a spring:
(Equation 1) PE = (1/2) * k * x^2
Where:
PE is the potential energy stored in the spring,
k is the force constant of the spring, and
x is the compression distance.
Initially, the block has potential energy given by:
(Equation 2) PE_initial = m * g * h
Where:
m is the mass of the block,
g is the acceleration due to gravity, and
h is the initial height of the block above the free end of the spring.
Since energy is conserved, the initial potential energy is equal to the final potential energy stored in the spring:
(Equation 3) PE_initial = PE_final
Combining equations 2 and 3:
m * g * h = (1/2) * k * x^2
Now we can substitute the given values into the equation and solve for x.
m = 1.60 kg
g = 9.8 m/s^2
h = 1.50 m
k = 1200 N/m
Plugging these values into the equation:
(1.60 kg) * (9.8 m/s^2) * (1.50 m) = (1/2) * (1200 N/m) * x^2
Simplifying the equation:
23.52 = 600 * x^2
Rearranging to solve for x:
x^2 = 23.52 / 600
x^2 = 0.0392
Taking the square root of both sides:
x = √0.0392
x ≈ 0.198 m
Therefore, the spring compresses by approximately 0.198 meters.