There is a rocket on a stage connected a spring with the spring constant k(N/m). The weights of the rocket and the stage are m(kg) and M(kg), respectively. The length of the spring without the stage and rocket or free is h0(m) and it becomes shrunk to h(m) on putting the rocket on it. Now, the rocket is launched with the initial velocity of v0. In this situation, answer the below questions, where the gravitational acceleration is g (m/s2), and the weight of the spring and resistance force due to air are negligible.
c) By launching, the spring is more shrunk. What is the energy stored in the spring when it is most shrunk?
d) What is the length of the spring at (c) situation.
To answer parts c) and d), we need to understand the concept of potential energy stored in a spring and apply the laws of conservation of energy.
c) The energy stored in the spring can be determined using the equation for potential energy stored in a spring, which is given by:
Potential energy stored in a spring (PE) = (1/2) * k * (h0 - h)^2
Where:
k is the spring constant (N/m)
h0 is the initial length of the spring without the rocket (m)
h is the length of the spring with the rocket (m)
In this case, the spring is most shrunk when the rocket is launched. Therefore, we need to find the value of h when the rocket is launched.
Since the spring is connected to the rocket and stage, the gravitational potential energy of the system decreases as the rocket is launched upward. This decrease in gravitational potential energy is transferred to the spring as potential energy stored in it.
The initial potential energy of the system can be calculated using the equation:
Initial Potential Energy (PE_initial) = (m + M) * g * h0
Where:
m is the mass of the rocket (kg)
M is the mass of the stage (kg)
g is the gravitational acceleration (m/s^2)
As the rocket is launched, the potential energy stored in the spring (PE) reaches its maximum value when the potential energy of the system becomes zero. At this point, all the initial potential energy is transferred to the spring.
Therefore, we can set up the equation:
PE + PE_initial = 0
Substituting the equation for potential energy stored in the spring and the initial potential energy, we get:
(1/2) * k * (h0 - h)^2 + (m + M) * g * h0 = 0
Simplifying the equation, we can solve for h:
(1/2) * k * (h0 - h)^2 = - (m + M) * g * h0
Now, we have an equation to find the value of h (the length of the spring when it is most shrunk) when the rocket is launched.
d) To find the length of the spring at situation (c), we simply substitute the value of h obtained from solving the equation above into the expression for h.
c) To find the energy stored in the spring when it is most shrunk, we can use the concept of potential energy stored in a spring. The potential energy stored in a spring is given by the formula:
PE = (1/2)kΔx^2
Where PE is the potential energy, k is the spring constant, and Δx is the displacement of the spring from its equilibrium position.
In this case, the initial length of the spring (without the rocket) is h0, and it becomes shrunk to h when the rocket is put on it. The displacement of the spring is Δx = h0 - h.
Therefore, the potential energy stored in the spring when it is most shrunk is given by:
PE = (1/2)k(h0 - h)^2
d) To find the length of the spring at the situation described in part (c), we need to consider the equilibrium position of the spring. When the rocket is launched, it will experience a force due to the spring, which can be balanced by the gravitational force acting on the rocket.
At equilibrium, the force exerted by the spring is equal in magnitude and opposite in direction to the weight of the rocket. Therefore, we can set up the following equation:
k(h0 - h) = m*g
Solving this equation for h, we get:
h = h0 - (m*g)/k
So, the length of the spring at the situation described in part (c) is h0 - (m*g)/k.