a spring is 30 cm long when a mass of 50 grams is placed when it is at rest. when an additional mass of 20 grams is placed, the spring elongates by 2.5 cm. what is the spring constant and what is the original length of the spring?
To find the spring constant and the original length 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.
Let's break down the problem step by step:
Step 1: Find the spring constant (k).
Hooke's Law can be represented as F = kx, where F is the force applied to the spring, k is the spring constant, and x is the displacement of the spring.
In the problem, the spring elongates by 2.5 cm when an additional mass of 20 grams is placed. We need to convert these values to the SI units: meters and newtons.
Mass of the additional weight: 20 grams = 0.02 kg (1 gram = 0.001 kg)
Displacement of the spring: 2.5 cm = 0.025 m (1 cm = 0.01 m)
Now, we can use Hooke's Law to find the spring constant (k).
F = kx
mg = kx (where m is the mass and g is the acceleration due to gravity)
Since the spring is at rest, the force applied by the spring is equal to the force exerted by the mass.
(0.02 kg)(9.8 m/s^2) = k(0.025 m)
0.196 N = k(0.025 m)
Therefore, the spring constant (k) is given by:
k = 0.196 N / 0.025 m = 7.84 N/m
Step 2: Find the original length of the spring.
To find the original length of the spring, we need to consider the initial mass and elongation of the spring.
In the problem, when a mass of 50 grams is placed on the spring, it is at rest. We need to find the elongation caused by this mass.
Using Hooke's Law, we can calculate the elongation:
F = kx
mg = kx
(0.05 kg)(9.8 m/s^2) = (7.84 N/m) x
0.49 N = (7.84 N/m) x
Therefore, the elongation of the spring caused by a mass of 50 grams is given by:
x = 0.49 N / 7.84 N/m = 0.0625 m = 6.25 cm
Since the elongation is the difference between the final length and the original length, we can write:
Final length = Original length + Elongation
Final length = Original length + 6.25 cm
Given that the spring elongates by 6.25 cm when a mass of 50 grams is placed, we can conclude that the original length of the spring is:
Original length = Final length - 6.25 cm
Original length = 30 cm - 6.25 cm
Original length = 23.75 cm
Therefore, the spring constant (k) is 7.84 N/m, and the original length of the spring is 23.75 cm.