There is a box attached to a spring. The spring has a spring constant of 39 N/m. The ì between the box and the surface is 0.3. The box has a mass of 1 kg.
What is the maximum distance that the spring can be stretched from equilibrium before the box begins to slide back?
What you call "The ì" must be the static friction coefficient, mus.
Let X be the maxium displacement without slipping.
k*X = M*g*mus
k is the spring constant.
You know what g and M are.
Solve for X
To find the maximum distance that the spring can be stretched from equilibrium before the box begins to slide back, we need to consider the forces acting on the box.
First, let's calculate the force exerted by the spring. The force exerted by a spring is given by Hooke's Law:
F = k * x
Where F is the force, k is the spring constant, and x is the displacement from equilibrium.
In this case, the spring constant is given as 39 N/m, and we want to find the maximum distance before the box begins to slide back, so the force exerted by the spring should be equal to the maximum static friction force.
Next, let's calculate the maximum static friction force. The maximum static friction force is given by:
F_friction = ì * N
Where ì is the coefficient of friction and N is the normal force. In this case, the normal force is equal to the weight of the box, which is given by:
N = m * g
Where m is the mass of the box (1 kg) and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Now, we can equate the force exerted by the spring with the maximum static friction force:
k * x = ì * m * g
Rearranging the equation to solve for x, we get:
x = (ì * m * g) / k
Substituting the given values:
x = (0.3 * 1 kg * 9.8 m/s^2) / 39 N/m
Simplifying the equation gives:
x = (2.94 N) / (39 N/m)
x = 0.075 m
Therefore, the maximum distance that the spring can be stretched from equilibrium before the box begins to slide back is 0.075 meters.