A 13 000-N car starts at rest and rolls down a hill from a height of 10.0 m (Fig. 13.6). It then moves across a level surface and collides with a light spring-loaded guardrail. (a) Neglecting any losses due to friction, find the
maximum distance the spring is compressed.
Assume a spring constant of 1.0 x 10^6 N/m.
(b) Calculate the maximum acceleration of the car after contact with the spring, assuming no frictional losses. (c) If the spring is compressed by only 0.30 m, find
the energy lost through friction.
To solve this problem, we can use the conservation of mechanical energy. The potential energy of the car at the top of the hill is converted into both kinetic energy as it rolls down the hill and potential energy stored in the compressed spring.
(a) First, let's find the maximum distance the spring is compressed. We can use the principle of conservation of mechanical energy:
Potential energy at the top of the hill = Kinetic energy at the bottom of the hill + Potential energy stored in the spring
mgh = (1/2)mv^2 + (1/2)kx^2
Where:
m = mass of the car = 13,000 N / 9.8 m/s^2 = 1326.5 kg
g = acceleration due to gravity = 9.8 m/s^2
h = height of the hill = 10.0 m
k = spring constant = 1.0 x 10^6 N/m
x = distance the spring is compressed (unknown)
Plugging in the given values, we have:
1326.5 kg * 9.8 m/s^2 * 10.0 m = (1/2) * 1326.5 kg * v^2 + (1/2) * (1.0 x 10^6 N/m) * x^2
130,041 J = 663.25 kg * v^2 + 0.5 x 10^6 N/m * x^2
At the maximum compression, the car momentarily comes to rest, so its final velocity is 0. Therefore, the equation becomes:
130,041 J = 663.25 kg * 0 + 0.5 x 10^6 N/m * x^2
130,041 J = 0.5 x 10^6 N/m * x^2
x^2 = (130,041 J) / (0.5 x 10^6 N/m) = 0.260082 m^2
x = sqrt(0.260082) = 0.5100 m
So, the maximum distance the spring is compressed is 0.510 m.
(b) To find the maximum acceleration of the car after contact with the spring, we can use Hooke's law:
F = k * x
Where F is the restoring force of the spring, which is also equal to the force exerted on the car due to the spring:
F = ma
So, we have:
ma = k * x
The mass of the car is 1326.5 kg and the spring constant is 1.0 x 10^6 N/m. Plugging in these values:
(1326.5 kg) * a = (1.0 x 10^6 N/m) * (0.510 m)
a = (1.0 x 10^6 N/m) * (0.510 m) / (1326.5 kg)
a = 382.23 m/s^2
So, the maximum acceleration of the car after contact with the spring is 382.23 m/s^2.
(c) Finally, to find the energy lost through friction when the spring is compressed by 0.30 m, we can use the principle of conservation of mechanical energy again:
Potential energy at the top of the hill = Kinetic energy at the bottom of the hill + Potential energy stored in the spring + Energy lost through friction
mgh = (1/2)mv^2 + (1/2)kx^2 + Efriction
We already know the values of m, g, h, k, and x from earlier calculations. Let's set up the equation:
1326.5 kg * 9.8 m/s^2 * 10.0 m = (1/2) * 1326.5 kg * v^2 + (1/2) * (1.0 x 10^6 N/m) * (0.30 m)^2 + Efriction
130,041 J = 663.25 kg * v^2 + (1/2) * (1.0 x 10^6 N/m) * (0.30 m)^2 + Efriction
130,041 J = 663.25 kg * 0 + (1/2) * (1.0 x 10^6 N/m) * (0.30 m)^2 + Efriction
Efriction = 130,041 J - (1/2) * (1.0 x 10^6 N/m) * (0.30 m)^2
Efriction = 130,041 J - 4500 J
Efriction = 125,541 J
So, the energy lost through friction when the spring is compressed by 0.30 m is 125,541 J.
To solve this problem, we can use the principle of conservation of mechanical energy. We will split the problem into three parts: the car rolling down the hill, the car moving across the level surface, and the car colliding with the spring-loaded guardrail.
(a) Finding the maximum distance the spring is compressed:
1. First, calculate the potential energy (PE) of the car at the top of the hill using the formula PE = mgh, where m is the mass of the car, g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the height of the hill (10.0 m). Since the car's weight is given as 13,000 N, divide this value by g to determine the mass of the car: m = 13,000 N / 9.8 m/s².
2. Calculate the potential energy at the top of the hill: PE = (mass of the car) × g × h.
3. Next, calculate the maximum kinetic energy (KE) of the car at the bottom of the hill using the formula KE = 0.5 × m × v², where v is the velocity of the car at the bottom of the hill.
4. Since the car starts from rest and rolls down the hill, the initial velocity (u) at the top of the hill is 0 m/s. The final velocity (v) at the bottom of the hill can be found using the equation v² = u² + 2as, where a is the acceleration due to gravity and s is the distance traveled down the hill (which is equal to the height of the hill, h). Solve for v.
5. Calculate the maximum kinetic energy at the bottom of the hill: KE = 0.5 × m × v².
6. To find the total mechanical energy at the bottom of the hill, add the potential and kinetic energy values obtained in steps 2 and 5: Total mechanical energy = PE + KE.
7. When the car collides with the spring-loaded guardrail, the mechanical energy is converted into elastic potential energy stored in the compressed spring. Use the formula for elastic potential energy: EPE = 0.5 × k × x², where k is the spring constant (1.0 × 10^6 N/m) and x is the displacement of the spring (to be determined).
8. Set the total mechanical energy at the bottom of the hill equal to the elastic potential energy stored in the compressed spring and solve for x: Total mechanical energy = EPE.
(b) Finding the maximum acceleration of the car after contact with the spring:
1. Use Hooke's law to find the force exerted by the compressed spring on the car: F = kx, where k is the spring constant and x is the displacement of the spring (0.30 m).
2. Convert this force into the acceleration of the car using Newton's second law: F = ma, where m is the mass of the car.
(c) Finding the energy lost through friction:
1. Since we are neglecting any losses due to friction, the energy lost through friction is assumed to be zero.
Note: Double-check your calculations and units to ensure accuracy.