A 5.00 kg carrying a charge of Q=60 uC is connected to a spring for which k=100N/m. The block lies on a frictionless horizontal track and the system is immersed in a uniform electric field of E=5.0e5 V/m. If the block is released from rest when the spring is unstretched.
A) By what maximum amount does the spring expand?
B) What is the equilibrium position of the block?
C) if the block is relesed from rest at x=0, how fast is it moving at x=0.25m?
D) now if instead of being released from rest the block is shoved to the left so that it starts with a velocity to the left of 2 m/s, what maximum distance to the left of x=0 would the block reach?
E) what maximum distance to the right of x=0 would the block reach?
F) where would the new equilibrium location be?
G) what if the block was shoved to the right giving it an initial velocity of 2 m/s. What maximum distance to the right of x=0 would the block reach?
To solve this problem, we need to consider the forces acting on the block, including the spring force and the electric force. We can use Newton's second law to determine the block's motion. Let's break down the questions one by one:
A) By what maximum amount does the spring expand?
To find the maximum amount the spring expands, we need to find the maximum displacement of the block from the equilibrium position. Hooke's Law states that the force exerted by a spring is proportional to its displacement, given by F = -kx, where k is the spring constant and x is the displacement. We can equate this force to the force due to the electric field, F = qE, where q is the charge and E is the electric field.
So, -kx = qE
-100x = (60 * 10^(-6)) * (5 * 10^5)
Now, we solve for x:
x = (60 * 10^(-6)) * (5 * 10^5) / 100
B) What is the equilibrium position of the block?
The equilibrium position of the block occurs when the net force acting on it is zero. From the previous equation, we have:
-kx = qE
Let's solve for x when the net force is zero:
0 = (60 * 10^(-6)) * (5 * 10^5) - kx
Solving for x:
x = (60 * 10^(-6)) * (5 * 10^5) / k
C) If the block is released from rest at x = 0, how fast is it moving at x = 0.25 m?
To determine the velocity of the block at x = 0.25 m, we can use energy conservation. The initial potential energy stored in the spring is converted into kinetic energy as the block moves. At the equilibrium position, all the potential energy is converted into kinetic energy. We can use the principle of conservation of mechanical energy to solve for the velocity.
D) Now, if instead of being released from rest, the block is shoved to the left with an initial velocity of 2 m/s. What maximum distance to the left of x = 0 would the block reach?
When the block is given an initial velocity, the problem becomes a case of simple harmonic motion with an initial displacement and velocity. We can use the equations of motion for simple harmonic motion to solve for the maximum distance reached by the block.
E) What maximum distance to the right of x = 0 would the block reach?
Similar to the previous case, the block will also reach a maximum distance to the right of x = 0. By applying the equations of motion for simple harmonic motion, we can solve for this distance.
F) Where would the new equilibrium location be?
The new equilibrium location is determined based on the net force acting on the block. By analyzing the forces, we can solve for the new equilibrium position.
G) What if the block was shoved to the right, giving it an initial velocity of 2 m/s? What maximum distance to the right of x = 0 would the block reach?
In this case, the block is given an initial velocity to the right. Again, we can use the equations of motion for simple harmonic motion to determine the maximum distance reached by the block.