If a masss of 65kg is dropped 4.3m onto a spring with a stiffness coefficient of 6.2x10^4 n/m, how far is the spring depressed?
M g (H +x) = (1/2)kx^2
Solve for x.
H = 4.3 m
k is the spring constant
M is the mass
Thanks
You're welcome.
They may want you to solve it without the x on the left side. That will give you nearly the same answer, and make solving easier, but it will be wrong. Gravity continues to act as the spring is compressed.
how much force is required to accelerate an 8.00 kg mass at 15.00 m/s2?
To find how far the spring is depressed, we can use Hooke's Law, which states that the force applied on a spring is directly proportional to its deformation.
Hooke's Law can be expressed as F = kx, where F is the force applied on the spring, k is the stiffness coefficient, and x is the displacement (depression) of the spring from its equilibrium position.
In this case, the mass of 65 kg is dropped onto the spring, so the force applied on the spring is equal to the weight of the mass, which is F = mg, where g is the acceleration due to gravity (approximately 9.8 m/s^2).
So, F = 65 kg x 9.8 m/s^2 = 637 N.
Using Hooke's Law, we can rearrange the formula to solve for x:
x = F / k
Substituting the values F = 637 N and k = 6.2 x 10^4 N/m into the formula:
x = 637 N / (6.2 x 10^4 N/m)
Now calculate this expression to find the depression of the spring.