A vertical spring with constant 830 N/m has a ball of mass 3 kg attached to the bottom of it, which is held with the spring unstretched.
(a) How far must the ball be lowered to reach its equilibrium position?
(b) The ball is given a charge of 0.060 C. If an electric field directed upward is applied, increasing slowly to a maximum value of 270.0 N/C, how far below the unstretched position is the new equilibrium position of the ball?
To find the answers to these questions, we can use Hooke's Law for the vertical spring and the restoring force equation for a charged particle in an electric field. Let's break down both parts of the problem:
(a) How far must the ball be lowered to reach its equilibrium position?
To determine the equilibrium position of the ball attached to the vertical spring, we need to first calculate the displacement or elongation of the spring.
Hooke's Law states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position. Mathematically, this can be expressed as:
F = -kx
Where F is the force exerted by the spring, k represents the spring constant, and x is the displacement or elongation of the spring from its equilibrium position.
Given:
k (spring constant) = 830 N/m
m (mass of the ball) = 3 kg
To determine the displacement (x), we can rearrange Hooke's Law as follows:
x = -F/k
Substituting the given values into the equation, we get:
x = -(m * g)/k
Where g is the acceleration due to gravity (approximately 9.8 m/s²).
Plugging in the values, we find:
x = -((3 kg) * (9.8 m/s²)) / (830 N/m)
Simplifying the expression:
x = -0.035 m
Therefore, the ball must be lowered by approximately 0.035 meters to reach its equilibrium position.
(b) How far below the unstretched position is the new equilibrium position of the ball?
For this part of the problem, we need to consider the electric force exerted on the charged ball in an upward electric field. The electric force can be calculated using the following equation:
F = qE
Where F is the force, q is the charge, and E is the electric field.
Given:
q (charge on the ball) = 0.060 C
E (electric field strength) = 270.0 N/C
To determine the displacement (d) below the unstretched position, we can calculate the gravitational force (Fg) and the electric force (Fe).
Gravitational force (Fg) can be given by:
Fg = mg
Where m represents the mass of the ball and g is the acceleration due to gravity.
Electric force (Fe) can be given by:
Fe = qE
Now, to find the displacement (d), we can compare the net force acting on the ball to the force exerted by the spring. At the equilibrium position, these two forces must be equal in magnitude but opposite in direction. The net force can be found by:
Net Force = Fe - Fg
At equilibrium, the magnitude of the force exerted by the spring (Fs) is equal to the net force.
Therefore,
Fs = Fe - Fg
Using the values:
Fs = k * d
Where k is the spring constant and d is the displacement of the spring from its equilibrium position.
Equating the forces:
k * d = Fe - Fg
Solving for d:
d = (Fe - Fg) / k
Substituting the given values into the equation:
d = (qE - mg) / k
Calculating the values:
d = ((0.060 C) * (270.0 N/C) - (3 kg) * (9.8 m/s²)) / (830 N/m)
Simplifying the expression:
d = 0.003 m
Therefore, the new equilibrium position of the ball, below the unstretched position, is approximately 0.003 meters.