A spring is compressed 12 cm. A 100-g block is placed against the spring. When the block is released, the block rises 2.0 m up the slop. Neglect friction. What is the spring constant of the spring?
To find the spring constant of the spring, we can use Hooke's Law. Hooke's Law 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:
F = k * x
Where:
F is the force exerted by the spring,
k is the spring constant, and
x is the displacement of the spring.
In this case, the spring is compressed by 12 cm, which is equal to 0.12 m. The block rises 2.0 m up the slope. Since the spring is compressed initially, we can take the displacement x as the sum of the initial compression and the distance the block rises:
x = 0.12 m + 2.0 m = 2.12 m
Next, we need to find the force exerted by the spring. Since the block goes up the slope, the force of gravity (weight) is directed opposite to the force exerted by the spring. Therefore, we can calculate the force exerted by the spring as the weight of the block:
F = m * g
Where:
m is the mass of the block, and
g is the acceleration due to gravity (9.8 m/s²).
To find the mass of the block, we convert the given weight of the block from grams to kilograms:
m = 100 g * (1 kg / 1000 g) = 0.1 kg
Now, we can calculate the force exerted by the spring:
F = 0.1 kg * 9.8 m/s² = 0.98 N
We can substitute the known values into the Hooke's Law equation and solve for the spring constant:
k * x = = F / x
k = 0.98 N / 2.12 m = 0.4623 N/m
Therefore, the spring constant of the spring is approximately 0.4623 N/m.