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? Please show your work.
Kinetic energy=final PE
1/2 m v^2=1/2 k x^2
solve for x
To solve this problem, we can use the principles of conservation of energy and Hooke's law.
First, let's calculate the initial kinetic energy of the car:
Kinetic energy (KE) = 1/2 * mass * velocity^2
Given:
Mass of the car (m) = 100 kg
Velocity of the car (v) = 22 m/s
KE = 1/2 * 100 kg * (22 m/s)^2
KE = 1/2 * 100 kg * 484 m^2/s^2
KE = 24200 kg*m^2/s^2
The initial kinetic energy of the car is 24200 kg*m^2/s^2.
Next, let's find the potential energy stored in the spring when it is compressed by a certain distance (x).
Potential energy (PE) stored in a spring = 1/2 * spring constant * displacement^2
Given:
Spring constant (k) = 4200 N/m
PE = 1/2 * 4200 N/m * x^2
PE = 2100 N/m * x^2
Since energy is conserved, the initial kinetic energy is equal to the potential energy stored in the spring when the car comes to a stop:
KE = PE
24200 kg*m^2/s^2 = 2100 N/m * x^2
Now, let's solve for x, the displacement of the spring:
x^2 = (24200 kg*m^2/s^2) / (2100 N/m)
x^2 = 11.5238 m
Taking the square root of both sides:
x = sqrt(11.5238 m)
x ≈ 3.39 m
Therefore, the car will displace the spring by approximately 3.39 meters before coming to a stop.