A 1100 kg car rolling on a horizontal surface has speed v = 70 km/h when it strikes a horizontal coiled spring and is brought to rest in a distance of 2.2 m. What is the spring stiffness constant of the spring?
The car comes to rest (temporarily)at compression distance X, at which point the initial kiteci energy (1/2) M V^2 is converted to spring potential energt, (1/2)kX^2.
Therefore
kX^2 = mV^2
k = m (V/X)^2
Make sure you convert V to meters per second before using the formula. k will be in Newtons/meter
To find the spring stiffness constant, we can use Hooke's Law, which states that the force exerted by a spring is proportional to the displacement of the spring from its equilibrium position.
The equation for Hooke's Law is:
F = -kx
Where:
F = Force exerted by the spring
k = Spring stiffness constant
x = Displacement of the spring from its equilibrium position
In this case, the car is brought to rest by the spring, so the force exerted by the spring will be equal to the force exerted by the car before it is stopped.
The force exerted by the car is given by Newton's second law of motion:
F = ma
Where:
m = Mass of the car
a = Acceleration of the car
Given:
Mass of the car, m = 1100 kg
Initial velocity of the car, v = 70 km/h
Final velocity of the car, 0 (since it comes to rest)
Initial displacement of the spring, x = 2.2 m
First, we need to convert the initial velocity from km/h to m/s:
v = 70 km/h = (70 * 1000) / 3600 m/s = 19.44 m/s
Next, we can calculate the acceleration of the car using the formula:
a = (vf - vi) / t
Where:
vf = Final velocity of the car = 0 m/s
vi = Initial velocity of the car = 19.44 m/s
t = Time taken to stop, which is not given
Since the car is brought to rest by the spring in a distance of 2.2 m, the time taken to stop can be found using the following equation of motion:
v^2 = u^2 + 2as
Where:
u = Initial velocity of the car = 19.44 m/s
v = Final velocity of the car = 0 m/s
s = Distance covered by the car = 2.2 m
Rearranging the equation, we get:
0 = (19.44)^2 + 2a(2.2)
Simplifying, we find:
2.2a = -19.44^2
a = (-19.44^2) / 2.2
Now, we can calculate the force exerted by the car using Newton's second law of motion:
F = ma
Substituting the values, we get:
F = 1100 * (-19.44^2) / 2.2
Finally, we can determine the spring stiffness constant, k, by rearranging Hooke's Law equation:
k = -F / x
Substituting the values, we find:
k = -[1100 * (-19.44^2) / 2.2] / 2.2
Calculating this expression will yield the spring stiffness constant.
To find the spring stiffness constant, also known as the spring constant or spring stiffness, we can use Hooke's Law. Hooke's Law states that the force exerted by a spring is proportional to the displacement of the spring from its equilibrium position. Mathematically, it can be expressed as:
F = k * x
Where:
- F is the force exerted by the spring,
- k is the spring stiffness constant,
- x is the displacement of the spring from its equilibrium position.
In this case, the car is brought to rest by the spring in a distance of 2.2 m. We can assume that the car compresses the spring by this amount, so x = 2.2 m.
Before the car comes to rest, it has an initial speed v = 70 km/h. We need to convert this to m/s as the SI unit of force is the Newton (N), not km/h. The conversion factor is 1 km/h = 0.2778 m/s. Therefore:
v = 70 km/h = 70 * 0.2778 m/s = 19.44 m/s
To find the spring stiffness constant, we need to calculate the force exerted by the spring. Since the car comes to rest, the total work done by the spring must equal the initial kinetic energy of the car:
Work = ΔKE
(½)mv² = kx²/2
Rearranging the equation, we get:
k = (½)mv²/x²
Plugging in the given values:
k = (0.5)(1100 kg)(19.44 m/s)² / (2.2 m)²
Calculating this expression will give us the spring stiffness constant.