when a mass of 2.5kg string is attached to a spring it stretches 42cm.
a) find the constant of the spring 'k'
b) how much force is required to stretch the spring 50cm?
c) how much energy is stored in the spring when stretched 0.35cm?
I guess it is hanging down on earth where g = 9.81 m/s^2 approximately
F = m g = k x
2.5 * 9.81 = k (.42)
solve for k
b) F = k (0.50)
c) U = (1/2) k x^2 = (1/2)(k)(0.35)^2
a) To find the constant of the spring (k), 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. The formula for Hooke's Law is given by:
F = k * x
Where F is the force exerted by the spring, k is the spring constant, and x is the displacement from the equilibrium position.
In this case, we know that the mass of the object attached to the spring is 2.5kg and the displacement is 42cm (or 0.42m). We can use this information to calculate the force (F) exerted by the spring.
First, we need to convert the displacement to meters:
x = 0.42m
Now, we can use Newton's second law to find the force:
F = m * g
Where m is the mass and g is the acceleration due to gravity (approximately 9.8 m/s^2).
F = 2.5kg * 9.8 m/s^2
F = 24.5 N
Now, equating this with the equation for Hooke's Law, we can solve for k:
24.5 N = k * 0.42m
k = 24.5 N / 0.42m
Therefore, the spring constant (k) is approximately 58.3 N/m.
b) To find the amount of force required to stretch the spring 50cm (or 0.5m), we can again use Hooke's Law:
F = k * x
where F is the force, k is the spring constant, and x is the displacement.
Plugging in the values, we have:
F = 58.3 N/m * 0.5m
F = 29.15 N
Therefore, the force required to stretch the spring 50cm is approximately 29.15 N.
c) To find the amount of energy stored in the spring when stretched 0.35cm (or 0.0035m), we can use the formula for potential energy stored in a spring:
E = (1/2) * k * x^2
Where E is the energy, k is the spring constant, and x is the displacement.
Plugging in the values, we have:
E = (1/2) * 58.3 N/m * (0.0035m)^2
E = (1/2) * 58.3 N/m * 0.00001225 m^2
E = 0.000035594875 J
Therefore, the energy stored in the spring when stretched 0.35cm is approximately 0.000035594875 Joules.