100 kg car moving at 22 m/s runs into a horizontal spring with spring constant 4200 N/m. How far does the car displace the spring before coming to a stop?
If you want expert help, you'll type the correct subject in the School Subject box.
Using conservation of anergy is the quickest way to do this.
It compresses the spring a distance X such that
(1/2) m V^2 = (1/2) k X^2
X = sqrt(m/k) * V = 3.4 m
It must be a very long spring.
thank you!
To solve this problem, we need to use the principle of conservation of mechanical energy. The initial kinetic energy of the car will transform into potential energy stored in the compressed spring.
The initial kinetic energy of the car is given by:
KE = (1/2) * mass * velocity^2
= (1/2) * 100 kg * (22 m/s)^2
= 24200 J
Since the car comes to a stop, all of the kinetic energy will be converted into potential energy stored in the spring. The potential energy stored in a spring is given by:
PE = (1/2) * k * x^2
where k is the spring constant and x is the displacement of the spring from its equilibrium position.
Let's rearrange the equation to solve for x:
x = sqrt(2 * PE / k)
Substituting the values, we have:
x = sqrt(2 * 24200 J / 4200 N/m)
= sqrt(115.24 m^2)
≈ 10.74 m
Therefore, the car displaces the spring by approximately 10.74 meters before coming to a stop.