if a 65 kg mass drops vertically onto a spring wit stiffness coefficient of 6.2x10 to the 4th power , how far does it depress the spring?
wouldn't it depend how far the drop was?
Sorry, the fall was 4.3m.
To find how far the spring is depressed when a mass is dropped onto it, we can use Hooke's Law, which states that the force exerted by the spring is directly proportional to the displacement.
Hooke's Law: F = -kx
Where:
F is the force exerted by the spring (in Newtons),
k is the stiffness coefficient of the spring (in Newtons per meter),
x is the displacement or compression of the spring (in meters).
In this case, the mass of the object dropped onto the spring is 65 kg. We need to calculate the displacement (x) of the spring.
To find the force exerted by the spring, we need to consider the weight of the object. The weight of an object is given by:
Weight = mass × gravity
Where gravity is approximately 9.8 meters per second squared (m/s^2).
Weight = 65 kg × 9.8 m/s^2
Next, we can equate the force exerted by the spring to the weight of the object and solve for x:
-kx = Weight
Plugging in the values:
-6.2 × 10^4 N/m × x = 65 kg × 9.8 m/s^2
Now we can solve for x:
x = (65 kg × 9.8 m/s^2) / (6.2 × 10^4 N/m)
Simply multiply and divide the values:
x = 637 N / 62000 N/m
Calculating this:
x ≈ 0.01027 meters
Therefore, the spring depresses approximately 0.01027 meters when the 65 kg mass drops onto it.