A 1200 kg car rolling on a horizontal surface has spped v= 65 km/h when it strikes a horizontal coiled spring and is brought to rest in a distance of 2.9 m. What is the spring stiffness constant of the spring?
To find the spring stiffness constant, also known as the spring constant or force constant (k), we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position.
The formula for Hooke's Law is:
F = -kx
Where:
F is the force exerted by the spring,
k is the spring stiffness constant, and
x is the displacement of the spring.
In this case, we know that the car comes to rest after striking the coiled spring, which means the force exerted by the spring will be equal to the initial momentum of the car.
The momentum of the car is given by:
p = mv
Where:
m is the mass of the car,
v is the velocity of the car.
Given:
Mass of the car (m) = 1200 kg
Velocity of the car (v) = 65 km/h = 18.06 m/s (converted to m/s)
First, let's calculate the initial momentum of the car:
p = mv
p = 1200 kg * 18.06 m/s
p = 21672 kg·m/s
As the car is brought to rest, the force exerted by the spring is equal to the initial momentum of the car (since there is no other external force acting on the car):
F = 21672 N (Force is equal to the initial momentum)
Using Hooke's Law, we have:
F = -kx
Rearranging the equation:
k = -F / x
Substituting the values:
k = -21672 N / 2.9 m
k ≈ -7485.52 N/m
Note: The negative sign indicates that the force exerted by the spring is opposite to the direction of the displacement.
So, the spring stiffness constant of the spring is approximately 7485.52 N/m.
To find the spring stiffness constant of the spring, we can use the principle of conservation of mechanical energy. The initial kinetic energy of the car is converted into the potential energy stored in the spring when it comes to rest.
First, let's convert the speed of the car from km/h to m/s. We know that 1 km/h = 0.2778 m/s. So, the speed of the car in m/s is:
v = 65 km/h * 0.2778 m/s = 18.056 m/s
Next, we can determine the initial kinetic energy of the car using the formula:
KE = (1/2) * m * v^2
where KE is the kinetic energy, m is the mass of the car, and v is the speed of the car. Plugging in the values, we get:
KE = (1/2) * 1200 kg * (18.056 m/s)^2
= 194631.744 J
Since the car comes to rest and all of its initial kinetic energy is transferred to potential energy stored in the spring, the potential energy of the spring can be calculated using the formula:
PE = (1/2) * k * x^2
where PE is the potential energy, k is the spring stiffness constant, and x is the distance the car is compressed or stretched from its equilibrium position. In this case, x is given as 2.9 m. So, we have:
PE = 194631.744 J = (1/2) * k * (2.9 m)^2
Rearranging the equation to solve for k:
k = (2 * PE) / (x^2)
= (2 * 194631.744 J) / (2.9 m)^2
= 134510 N/m
Therefore, the spring stiffness constant of the spring is 134510 N/m.
This problem can be solved with the conservation of energy. The total amount of energy in the car is the same before and after the collision with the spring. The car's kinetic energy before the collision is completely converted to spring potential energy after the collision.
1/2 * m * v^2 = 1/2 * k * x^2
Plug and chug, solve for k