The spring constant for a vertical spring stretch 15cm when 22.0kg mass is attached to it is about:
a. 1440 N/m
b. 1210 N/m
c. 990 N/m
d. 17.1 N/m
e. 2.54 N/m
M*g = 22 * 9.8 = 215.6 N. = Wt. of the Mass.
k = 215.6N/0.15m = 1437.33N/m.
To find the spring constant, also known as the spring stiffness or force constant, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position.
Hooke's Law can be expressed mathematically as:
F = k * x
where F is the force applied to the spring, k is the spring constant, and x is the displacement from the equilibrium position.
In this case, we are given that the spring stretches by 15 cm (0.15 m) when a 22.0 kg mass is attached. We can assume that the displacement corresponds to the gravitational force acting on the mass.
So, we can use the equation:
F = m * g
where F is the force, m is the mass, and g is the acceleration due to gravity.
Plugging in the values:
F = (22.0 kg) * (9.8 m/s²) = 215.6 N
Now, we can rearrange Hooke's Law to solve for k:
k = F / x
Plugging in the values:
k = 215.6 N / 0.15 m = 1437.33 N/m
Rounding the answer to the nearest whole number, the spring constant is approximately 1440 N/m, which is option a.