A 0.50kg mass is suspended on a spring that stretches 3.0cm.
a. What is the spring constant?
b. What added mass would stretch the spring an additional 2.0cm?
c. What is the change in potential energy when the mass is added?
Conservation of energy is not my favorite......
(0.50*9.8N)/3cm = 1.63 N/cm
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To answer these questions, we need to understand Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position. The formula for Hooke's Law is F = -kx, where F is the force, k is the spring constant, and x is the displacement.
a. To find the spring constant, we can use Hooke's Law and the given information. Since the mass is at equilibrium, the force exerted by the spring is equal in magnitude but opposite in direction to the force due to gravity. So, we have F = mg, where m is the mass and g is the acceleration due to gravity. The force exerted by the spring is also given by F = kx. Equating the two equations, we have mg = kx. Rearranging the equation to solve for k, we get k = mg/x.
Given:
m = 0.50 kg
x = 3.0 cm = 0.03 m
Using the values, k = (0.50 kg)(9.8 m/s^2) / 0.03 m.
b. To find the added mass that would stretch the spring an additional 2.0 cm, we need to calculate the new displacement and apply Hooke's Law again. The total displacement is the initial displacement plus the added displacement, so the new total displacement is 0.03 m + 0.02 m = 0.05 m.
Using F = -kx, we can rearrange the equation to solve for the added mass, m. Rearranging, m = -k(x2 - x1) / g, where x1 is the initial displacement and x2 is the final displacement.
Given:
x1 = 0.03 m
x2 = 0.05 m
k = (0.50 kg)(9.8 m/s^2) / 0.03 m
g = 9.8 m/s^2
Substituting the values, m = -[(0.50 kg)(9.8 m/s^2) / 0.03 m][(0.05 m - 0.03 m) / (9.8 m/s^2)].
c. To find the change in potential energy when the mass is added, we can use the formula for potential energy stored in a spring, which is given by U = (1/2)kx^2. The initial potential energy of the spring is (1/2)kx₁², and the final potential energy is (1/2)kx₂².
Using the given values from part a, the initial potential energy is (1/2)(0.5 kg)(9.8 m/s²)(0.03 m)².
Using the values from part b, the final potential energy is (1/2)[(0.50 kg)(9.8 m/s²) / 0.03 m][0.05 m]².
The change in potential energy is the final potential energy minus the initial potential energy.