A 1370 kg car rolling on a horizontal surface has a speed of 20 km/hr when it strikes a horizontal coiled spring and is brought to rest in a distance of 5.5 m. What is the spring constant (in N/m) of the spring? (Ignore nonconservative forces such as friction.)
Loss in KE = Gain in elastic PE of the spring.
(1/2)mV^2 = (1/2)kx^2
find K from this equation. Remember to convert the units of V to m/s
To find the spring constant, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to its displacement from equilibrium. The formula for Hooke's Law is:
F = -kx
where F is the force exerted by the spring, k is the spring constant, and x is the displacement from equilibrium.
In this case, the car is brought to rest by the spring, so the force exerted by the spring is equal to the force applied by the car to decelerate. The force applied by the car can be calculated using Newton's second law:
F = ma
where F is the force, m is the mass, and a is the acceleration.
Given:
Mass of the car (m) = 1370 kg
Speed of the car (v) = 20 km/hr = 20 * 1000 m/3600 s = 5.56 m/s
Distance traveled by the car before coming to rest (x) = 5.5 m
To find the force applied by the car, we need to calculate the deceleration using the kinematic equation:
v^2 = u^2 + 2as
where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement.
Rearranging the equation:
a = (v^2 - u^2) / (2s)
Substituting the given values:
a = (0 - (5.56)^2) / (2 * 5.5)
Simplifying:
a = -30.976 m/s^2
Now that we have the acceleration, we can calculate the force applied by the car:
F = ma
F = 1370 kg * (-30.976 m/s^2)
F = -42537.152 N (negative sign indicates that the force is in the opposite direction of motion)
According to Hooke's Law, the force exerted by the spring is equal to the force applied by the car:
F = -kx
-42537.152 N = -k * 5.5 m
Solving for k:
k = -42537.152 N / 5.5 m
k ≈ 7725.75 N/m
Therefore, the spring constant of the spring is approximately 7725.75 N/m.