A body of mass 500g suspended from the end of a spiral spring which obeys Hooke's law, produced an extension of 10cm. If the mass is pulled down a distance of 50cm and released, calculate
a) the force constant of the spring
mass = 0.500 kg
Force = m g = 0.5 * 9.81 = 4.905 Newtons
extension = 0.10 meters
so
k = 4.905 / 0.1 = 49.05 Newtons / meter
I do not see any further questions but anyway
additional PE stored in spring if pulled further 0.50 meter
Potential Energy in spring = (1/2) k x^2 = (1/2) 49.05 (.25)
= 6.13 Joules
that will be the KE as it passes through equilibrium static stretchpoint
(1/2) m v^2 = .5 * .5 * v^2 = 6.13
so max v = 4.95 m/s
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To solve for the force constant of the spring, we can use Hooke's law equation: F = kx, where F is the force, k is the force constant, and x is the displacement.
In this case, we are given that the mass is 500g, which is equal to 0.5kg, and produces an extension of 10cm, which is equal to 0.1m.
We can calculate the force by multiplying the mass by the acceleration due to gravity (9.8 m/s^2). So, F = 0.5kg * 9.8 m/s^2 = 4.9 N.
Plugging in these values into Hooke's law equation, we have:
4.9 N = k * 0.1m.
To calculate the force constant k, we can rearrange the equation:
k = 4.9 N / 0.1m = 49 N/m.
Therefore, the force constant of the spring is 49 N/m.
To calculate the force constant of the spring, 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 as:
F = -kx
Where:
F is the force exerted by the spring,
k is the force constant (also known as the spring constant),
x is the displacement of the spring from its equilibrium position.
In this case, we are given that the spring extends by 10 cm (or 0.1 m) when a mass of 500 g (or 0.5 kg) is suspended from it. By substituting these values into Hooke's Law, we can solve for the force constant:
F = -kx
F = -0.5 kg * 9.8 m/s^2 (acceleration due to gravity) [Note: The negative sign indicates that the force is acting in the opposite direction of the displacement.]
0.5 kg * 9.8 m/s^2 = k * 0.1 m
Therefore, the force constant of the spring is:
k = (0.5 kg * 9.8 m/s^2) / 0.1 m
= 49 N/m
So, the force constant of the spring is 49 N/m.