A 1100-kg car moving on a horizontal surface has speed v = 60km/h when it strikes a horizontal coiled spring and is brought to rest in a distance of 3.1m .What is the spring stiffness constant of the spring?
Why did the car go to the spring? Because it wanted to have a "spring fling"! Now let's bounce right into the question.
To find the spring stiffness constant (k), we can use the equation for the work done by the spring:
Work = (1/2)kx^2
Here, x is the distance the spring compresses (3.1 m) and the work done by the spring is equal to the initial kinetic energy of the car. So we can set up the equation:
(1/2)kx^2 = (1/2)mv^2
Substituting the given values, we have:
(1/2)k(3.1)^2 = (1/2)(1100)(60/3.6)^2
Simplifying further:
4.95k = 1100(100/3.6)^2
Calculating the right side:
4.95k ≈ 847,222.22
Finally, divide both sides by 4.95 to find the spring stiffness constant:
k ≈ 170,929.50 N/m
So, the spring stiffness constant is approximately 170,929.50 N/m. Keep on springing those questions at me!
To find the spring stiffness constant, also known as the spring constant (k), we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to its displacement.
Hooke's Law can be written as:
F = -kx
where F is the force exerted by the spring, k is the spring constant, and x is the displacement of the spring.
In this case, the car is brought to rest, so the force exerted by the spring is equal to the force exerted by the car. We can calculate the force exerted by the car using Newton's second law:
F = ma
where F is the force, m is the mass of the car, and a is the acceleration.
Given:
Mass of the car (m) = 1100 kg
Initial speed (v) = 60 km/h = 16.67 m/s
Distance traveled (x) = 3.1 m
The force exerted by the car can be calculated as follows:
F = ma
F = m * (change in velocity / change in time)
The change in velocity is given by v - 0, since the car comes to rest:
F = m * (-v / t)
We can relate the distance traveled (x) and the time taken (t) using the formula:
x = v * t + (1/2) * a * t^2
Since the car comes to rest, v = 0:
x = (1/2) * a * t^2
We can rearrange this equation to solve for time (t):
t = sqrt((2 * x) / a)
Substituting this value of t into the equation for force, we get:
F = m * (-v / (sqrt((2 * x) / a)))
Now we can use Hooke's Law to find the spring constant (k):
F = -kx
-m * (v / (sqrt((2 * x) / a))) = -k * x
Simplifying the equation, we get:
k = m * (v^2) / (2 * x)
Now we can substitute the known values:
k = 1100 kg * (16.67 m/s)^2 / (2 * 3.1 m)
Calculating this, we find:
k ≈ 61683.87 N/m
Therefore, the spring stiffness constant of the spring is approximately 61683.87 N/m.
To find the spring stiffness constant, also known as the spring constant or force constant (k), we can use Hooke's Law. Hooke's Law states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position.
The equation 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 of the spring from its equilibrium position.
In this case, we want to find the spring constant (k). The car strikes the spring and is brought to rest, so the initial velocity (vi) is 60 km/h and the final velocity (vf) is 0. The displacement (x) of the spring is given as 3.1 m.
To start solving this problem, we need to convert the car's initial velocity from km/h to m/s.
1 km = 1000 m
1 h = 3600 s
So, to convert 60 km/h to m/s, we can use the conversion factor:
60 km/h * (1000 m / 1 km) * (1 h / 3600 s) ≈ 16.67 m/s
Now we have all the necessary values to find the spring constant.
We know that the initial kinetic energy (KEi) of the car is equal to the work done by the spring to bring the car to rest.
KEi = 1/2 * mv^2
Work = F * x
Since the work done by the spring is equal to the initial kinetic energy, we can set up the equation:
1/2 * mv^2 = F * x
We need to find the force (F) exerted by the spring. To do this, we can use Newton's second law of motion, which relates force, mass, and acceleration (a):
F = ma
The acceleration can be calculated using the following equation:
vf^2 = vi^2 + 2a(x - xi)
Since the initial velocity (vi) is 16.67 m/s, the final velocity (vf) is 0, and the initial displacement (xi) is 0, the equation simplifies to:
0 = (16.67 m/s)^2 + 2a(3.1 m)
Now we can solve for the acceleration (a):
2a(3.1 m) = - (16.67 m/s)^2
a = - (16.67 m/s)^2 / (2 * 3.1 m)
Once we have the value of acceleration, we can substitute it back into the equation F = ma to find the force (F).
Finally, we can substitute the force (F) and the displacement (x) into the equation for Hooke's Law to solve for the spring constant (k):
F = -kx
k = -F / x
By following these steps, we can find the spring stiffness constant (k).