Hooke's law states that the distance an object is stretched varied directly with the force exerted on the object. Suppose the force exerted by 7-kg mass stretches a certain spring 5 cm. To the nearest 0.1 cm. how far will the spring be stretched by the force of an 11-kg mass? please show step by step
To solve this problem using Hooke's Law, we need the formula:
F = kx
Where:
F = force exerted on the object
k = spring constant
x = distance the object is stretched
Step 1: Find the spring constant (k)
Since we are given the force exerted by a 7-kg mass (F = 7 kg) and the distance stretched (x = 5 cm), we can rearrange the formula:
F = kx
Solving for k:
k = F / x
k = 7 kg / 5 cm
k = 1.4 kg/cm
Step 2: Find the distance the spring will be stretched for an 11-kg mass (x)
Using the same formula, we can now find x:
F = kx
Rearranging for x:
x = F / k
x = 11 kg / 1.4 kg/cm
x = 7.86 cm
To the nearest 0.1 cm, the distance the spring will be stretched by the force of an 11-kg mass is approximately 7.9 cm.
To solve this problem using Hooke's Law, we can express it as an equation:
F = k * x
Where:
F - Force applied to the spring
k - Spring constant (a measure of stiffness of the spring)
x - Distance the spring is stretched or compressed
Now, we need to find the value of k (spring constant). We can obtain this value from the given information that a 7-kg mass stretches the spring by 5 cm.
Let's rearrange Hooke's law equation to solve for k:
k = F / x
We know that the force (F) is the weight of an object, which is equal to mass times gravitational acceleration (g ≈ 9.8 m/s²).
F = m * g
So, for a 7-kg mass:
F = 7 kg * 9.8 m/s² = 68.6 N
Substituting the values of F and x into the equation for k:
k = 68.6 N / 0.05 m = 1372 N/m
Now that we have the spring constant (k), we can calculate the distance the spring will be stretched by the force of an 11-kg mass.
Using Hooke's law equation:
F = k * x
We know that F = m * g, where m = 11 kg and g ≈ 9.8 m/s².
So, for an 11-kg mass:
F = 11 kg * 9.8 m/s² = 107.8 N
Rearranging the equation:
x = F / k
Calculating the value:
x = 107.8 N / 1372 N/m = 0.0788 m
Converting the answer to centimeters (cm):
x = 0.0788 m * 100 cm/m = 7.88 cm (rounded to the nearest 0.1 cm)
Therefore, the spring will be stretched approximately 7.9 cm (to the nearest 0.1 cm) by the force of an 11-kg mass.