A mass of 1.00kg is dropped from a height of 80cm onto a spring, of unstretched length of 20.0cm, which is resting on the floor. It comes to rest, momentarily, at a height of 10.0cm from the floor. If we can ignore friction, find the spring constant of the spring.
I feel like spring questions are so confusing. I'm not even sure what to do. I calculated -Wgravity=mgDELTAy= 7.84J
Now, can I equate -Wgravity=Uelastic=1/2kx^2?
I'm so confused.
Stretching the spring 10 cm casues it to have a spring potential energy equal to the gravitational potential energy loss.
(1/2) k*(0.10m)^2 = M g H = 1.0*9.8*0.7
= 6.86 J
Solve for k
0.7 m (70 cm) is the actual distance it has fallen at maximum compression.
0.10 m is the spring comoression.
Don't worry, I'm here to help you understand spring problems!
To find the spring constant of the spring, we can equate the gravitational potential energy (W_gravity) to the elastic potential energy (U_elastic).
First, let's calculate the gravitational potential energy:
W_gravity = m * g * Δy
Where m is the mass (1.00 kg), g is the acceleration due to gravity (9.81 m/s²), and Δy is the change in height (from 80 cm to 10 cm = 0.70 m).
W_gravity = 1.00 kg * 9.81 m/s² * 0.70 m
W_gravity = 6.867 J
Now, let's equate this to the elastic potential energy:
U_elastic = 1/2 * k * x²
Where k is the spring constant (which we want to find) and x is the displacement of the spring (from the unstretched length of 20.0 cm to the position where it comes to rest at 10.0 cm above the floor = 30.0 cm = 0.30 m).
Equating the two energies, we have:
6.867 J = 1/2 * k * (0.30 m)²
Now, let's solve for k:
k = (2 * 6.867 J) / (0.30 m)²
k = 38.15 N/m
So, the spring constant of the spring is approximately 38.15 N/m.
Remember, in these types of problems, it's important to consider the conservation of energy and equate the different forms of energy involved. Keep practicing and you'll get more comfortable with spring problems!