A 2000kg car begins from rest and rolls 50.0m down a frictionless 10 degrees incline. If there is a horizontal spring at the end of the incline, what spring constant is required to stop the car in a distance of 1.00m?
The car has lost potential energy the spring must absorb.
So the change height is 51sin10, and the mgh change of PE is then
2000*9.8*51sin10. But the trick here is stopping the car. THe energy must be absorbed by the spring, spring energy is 1/2 k 1^1, set them equal and
1/2 k=2000*9.8*51sin10 solve for k.
k=2*2000*9.8*51sin10
A book is placed on a horizontal wooden plane that is undergoing simple
harmonic motion with an amplitude of 1.0 m. The coefficient of friction
between the book and the horizontal wooden plane is given by micro = 0.5.
Determine the frequency of the horizontal wooden platform when the book
is about to slip from the horizontal wooden plane.
To find the spring constant required to stop the car in a distance of 1.00m, we need to determine the potential energy of the car on the inclined plane and equate it to the potential energy stored in the spring.
Here's how we can solve it step by step:
Step 1: Calculate the potential energy of the car on the inclined plane
The potential energy of an object on an inclined plane is given by the formula:
Potential energy = m * g * h
where m is the mass of the car, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the vertical height of the incline.
In this case, the mass of the car is 2000 kg, the acceleration due to gravity is 9.8 m/s^2, and the vertical height of the incline is given by the formula:
h = l * sin(θ)
where l is the length of the inclined plane (50.0m) and θ is the angle of the incline (10 degrees).
Plugging in the values, we have:
h = 50.0m * sin(10 degrees)
h ≈ 8.68m
Potential energy = 2000 kg * 9.8 m/s^2 * 8.68m
Potential energy ≈ 170,782 J
Step 2: Calculate the potential energy stored in the spring
The potential energy stored in a spring is given by the formula:
Potential energy = (1/2) * k * x^2
where k is the spring constant and x is the distance the spring is compressed.
In this case, we want the car to stop in a distance of 1.00m, so x = 1.00m.
Plugging in the values, we have:
Potential energy = (1/2) * k * (1.00m)^2
Potential energy = 0.5 * k * 1.00m^2
Potential energy = 0.5 * k * 1.00m^2
Potential energy = 0.5 * k J
Step 3: Equate the potential energies and solve for k
Equate the potential energy of the car on the inclined plane to the potential energy stored in the spring:
Potential energy of car = Potential energy of spring
170,782 J = 0.5 * k J
Divide both sides of the equation by 0.5:
341,564 J = k J
So, the required spring constant to stop the car in a distance of 1.00m is approximately 341,564 N/m.