The stored potential energy in the trampoline at maximum compression, (1/2) k X^2, equals the gravitational potential energy at maximum height, M g H. (Kinetic energy is zero in both cases)
X = 0.5 m
g = 9.8 m/s^2
H = 2.5 m
k = the spring constant.
Solve for k.
k = 2MgH/X^2
A 115 kg man is jumping on a trampoline. If the trampoline is compressed .5 meters and the maximum height is 2.5 meters above the trampoline, what is the spring constant?
A Knight of the Round Table fires off a vat of burning pitch from his catapult at 14.5 m/s, at 33 ◦ above the horizontal.
The acceleration of gravity is 9.8 m/s2 . How long is it in the air?
Answer in units of s
10.47
To solve for the spring constant (k), we can use the equation you provided:
k = 2MgH / X^2
Given values:
X = 0.5 m (compression of the trampoline)
g = 9.8 m/s^2 (acceleration due to gravity)
H = 2.5 m (maximum height above the trampoline)
M = 115 kg (mass of the man)
Let's substitute these values into the equation:
k = 2(115 kg)(9.8 m/s^2)(2.5 m) / (0.5 m)^2
Simplifying:
k = 2(115)(9.8)(2.5) / (0.5)^2
k = (2 * 115 * 9.8 * 2.5) / (0.25)
k = (5,332) / (0.25)
k = 21,328
Therefore, the spring constant (k) is equal to 21,328 N/m.