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?
7/5 = x
(x)(11) = ?
To solve this problem, we can use Hooke's law formula:
F = kx
Where:
F is the force exerted on the object
k is the spring constant
x is the distance the object is stretched or compressed
We are given that a 7-kg mass stretches the spring 5 cm. We can use this information to find the spring constant (k).
First, convert the mass to weight by multiplying it by the acceleration due to gravity (g):
Weight = mass * gravity
Weight = 7 kg * 9.8 m/s^2
Weight = 68.6 N
Since the weight is equal to the force exerted on the spring, we have:
68.6 N = k * 5 cm
Now we can solve for the spring constant (k):
k = 68.6 N / 5 cm
k = 13.72 N/cm
Now that we have the spring constant, we can use it to find how far the spring will be stretched by an 11-kg mass.
First, find the weight of the 11-kg mass:
Weight = mass * gravity
Weight = 11 kg * 9.8 m/s^2
Weight = 107.8 N
Now, we can use the spring constant to find the distance the spring will be stretched:
F = kx
107.8 N = 13.72 N/cm * x
Solving for x:
x = 107.8 N / 13.72 N/cm
x ≈ 7.862 cm
Therefore, the spring will be stretched by approximately 7.862 cm by the force of an 11-kg mass.