A 1.80 kg object is connected to a spring of force constant 120 N/m. How far is the spring stretched if it is used to drag the object across a floor at constant velocity? Assume the coefficient of kinetic friction is 0.60.
To find the distance by which the spring is stretched, we need to consider the forces acting on the object when it is being dragged across the floor at a constant velocity.
The force required to overcome the kinetic friction is given by the equation:
Ffriction = μ * N
where Ffriction is the force of kinetic friction, μ is the coefficient of kinetic friction, and N is the normal force.
The normal force is equal to the weight of the object, which is given by:
N = m * g
where m is the mass of the object and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Therefore, the force of kinetic friction can be expressed as:
Ffriction = μ * m * g
In this case, Ffriction is equal to the force exerted by the spring, which can be calculated using Hooke's law:
Fspring = k * x
where Fspring is the force exerted by the spring, k is the force constant of the spring, and x is the displacement or stretch of the spring from its equilibrium position.
Since the spring force is equal to the force of kinetic friction, we can equate the two equations and solve for x:
k * x = μ * m * g
Substituting the given values:
120 * x = 0.60 * 1.80 * 9.8
Simplifying the equation:
120 * x = 10.584
Now, solve for x by dividing both sides of the equation by 120:
x = 10.584 / 120
x ≈ 0.0882 meters
Therefore, the spring is stretched by approximately 0.0882 meters when used to drag the object across the floor at a constant velocity.