A 11.0kg weather rocket generates a thrust of 230N . The rocket, pointing upward, is clamped to the top of a vertical spring. The bottom of the spring, whose spring constant is 510 N/m , is anchored to the ground. After the engine is ignited, what is the rocket's speed when the spring has stretched 40.0cm ?
To find the rocket's speed when the spring has stretched 40.0 cm, we can use the principle of conservation of energy.
First, let's calculate the potential energy stored in the spring when it is stretched 40.0 cm. The potential energy stored in a spring is given by the equation:
PE = (1/2) k x^2
where PE is the potential energy, k is the spring constant, and x is the displacement of the spring from its equilibrium position.
Given:
k = 510 N/m (spring constant)
x = 40.0 cm = 0.40 m (displacement)
Plugging in the values, we can calculate PE:
PE = (1/2) * 510 N/m * (0.40 m)^2
= 20.4 J
Now, since the rocket is generating a thrust of 230 N, the work done by the thrust to stretch the spring can be calculated as:
Work = Force × distance
Work = 230 N × 0.40 m
= 92 J
According to the principle of conservation of energy, the work done by the thrust must equal the change in potential energy stored in the spring. Therefore, the change in potential energy (ΔPE) is given by:
ΔPE = Work
ΔPE = 92 J
Now, let's find the kinetic energy (KE) of the rocket when the spring has stretched 40.0 cm. The kinetic energy is given by the equation:
KE = (1/2) mv^2
where KE is the kinetic energy, m is the mass of the rocket, and v is the velocity (speed) of the rocket.
Given:
m = 11.0 kg (mass of the rocket)
ΔPE = 92 J (change in potential energy)
Using the conservation of energy, we can equate the change in potential energy to the change in kinetic energy:
ΔPE = ΔKE
92 J = (1/2) * 11.0 kg * v^2
Now, let's solve for v:
v^2 = (2 * 92 J) / 11.0 kg
v^2 = 16.727
v ≈ 4.09 m/s
Therefore, the rocket's speed when the spring has stretched 40.0 cm is approximately 4.09 m/s.