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.
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.
1 answer
1 answer