When a 2.00-kg object is hung vertically on a certain light spring described by Hooke's law, the spring stretches 3.17 cm.
(a) What is the force constant of the spring?
N/m
(b) If the 2.00-kg object is removed, how far will the spring stretch if a 1.00-kg block is hung on it?
. cm
(c) How much work must an external agent do to stretch the same spring 8.50 cm from its unstretched position?
(a) To find 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 from its equilibrium position.
Hooke's law can be written as: F = k * x
where F is the force applied, k is the force constant of the spring, and x is the displacement from the equilibrium position.
In this case, we have an object of mass 2.00 kg hanging vertically, causing the spring to stretch 3.17 cm. We can convert the mass to weight using the acceleration due to gravity (9.8 m/s^2):
Weight = mass * acceleration due to gravity
Weight = 2.00 kg * 9.8 m/s^2
The weight of the object is the force applied to the spring, so we can substitute this into Hooke's law equation:
Weight = k * x
2.00 kg * 9.8 m/s^2 = k * 0.0317 m
Now, we can solve for the force constant:
k = (2.00 kg * 9.8 m/s^2) / 0.0317 m
k ≈ 61.6 N/m
Therefore, the force constant of the spring is approximately 61.6 N/m.
(b) To calculate how far the spring will stretch when a 1.00-kg block is hung on it, we can again use Hooke's law.
Let's denote the displacement as x2 (stretch when 1.00-kg block is hung).
The weight of the 1.00-kg block is 1.00 kg * 9.8 m/s^2, which is the force applied to the spring. So we have:
Force = k * x2
1.00 kg * 9.8 m/s^2 = 61.6 N/m * x2
Now we can solve for x2:
x2 = (1.00 kg * 9.8 m/s^2) / 61.6 N/m
x2 ≈ 0.159 m
Multiplying by 100 to convert to cm, we get:
x2 ≈ 15.9 cm
Therefore, if a 1.00-kg block is hung on the spring, it will stretch approximately 15.9 cm.
(c) To calculate the work required to stretch the spring 8.50 cm from its unstretched position, we can use the formula for the work done by a spring:
Work = (1/2) * k * (x^2 - x0^2)
In this equation, k is the force constant of the spring, x is the final displacement, and x0 is the initial displacement (which is zero in this case since it's from the unstretched position).
We know the force constant of the spring is 61.6 N/m, and the final displacement is 8.50 cm (converting to meters: 0.085 m), and the initial displacement is 0.
Work = (1/2) * (61.6 N/m) * ((0.085 m)^2 - (0 m)^2)
Work ≈ 2.32 J
Therefore, the external agent must do approximately 2.32 Joules of work to stretch the spring 8.50 cm from its unstretched position.