A 950-kg car strikes a huge spring at a speed of 25 m/s, compressing the spring 4.0 m. (a) What is the spring stiffness constant of the spring? (b) How long is the car in contact with the spring before bouncing off in the opposite direction?
a)
x=4m
m=950kg
v=25m/s
1/2(m)(v)^2=1/2(k)(x)^2
^use this equation to solve for k, should get 37000 N/m
b)
T=2pi*sqrt(m/k)
Use this to solve for T, should get 1s.
T is the time period for the spring to go from equilibrium, to compressed, to stretched, and back to equilibrium. So basically the question is asking for half of T. So your answer should be 0.5s.
To find the spring stiffness constant, we can use Hooke's Law:
F = k * x
where:
F is the force exerted by the spring,
k is the spring stiffness constant, and
x is the displacement of the spring.
In this case, the car strikes the spring, causing it to compress. The force exerted by the spring is equal to the force exerted by the car. We can use Newton's second law of motion to find the force:
F = m * a
where:
m is the mass of the car, and
a is the deceleration of the car.
In this case, the car is brought to rest by the spring, so the deceleration is equal to its initial velocity squared divided by twice the displacement of the spring:
a = (v^2) / (2 * x)
Substituting this back into Newton's second law, we get:
F = m * ((v^2) / (2 * x))
Now we can equate this to Hooke's Law:
k * x = m * ((v^2) / (2 * x))
Simplifying, we find:
k = (m * (v^2)) / (2 * x^2)
(a) Plugging in the given values:
m = 950 kg, v = 25 m/s, x = 4.0 m
k = (950 * (25^2)) / (2 * (4.0^2))
Now we can calculate k.
(b) To find the duration of contact between the car and the spring, we need to calculate the time it takes for the car to compress the spring and then rebound to its original position. We can use the equations of motion for uniformly decelerated motion.
The final velocity of the car when it comes to rest can be calculated using the equation:
v^2 = u^2 - 2 * a * s
where:
u is the initial velocity of the car,
a is the deceleration, and
s is the displacement.
In this case, the initial velocity is 25 m/s, the final velocity is 0 m/s, the deceleration is given by a = (v^2) / (2 * x), and the displacement is x.
Substituting the values, we find:
0^2 = 25^2 - 2 * ((25^2) / (2 * 4.0)) * 4.0
Solving for u, we get:
u = sqrt(25^2 - 2 * ((25^2) / (2 * 4.0)) * 4.0)
Now we can find the time using the equation:
t = (v - u) / a
where v is the final velocity (0 m/s), u is the initial velocity, and a is the deceleration.
Plugging in the values, we can calculate the time for contact.
To solve part (a) of the problem, we can use the equation for the potential energy of a spring:
Elastic potential energy (PE) = 1/2 * k * x^2
Where:
PE is the potential energy stored in the spring (Joules)
k is the spring stiffness constant (N/m)
x is the compression or extension of the spring (m)
Given information:
Mass of the car (m) = 950 kg
Initial speed of the car (v) = 25 m/s
Compression of the spring (x) = 4.0 m
Since the car is coming to a stop and bouncing off in the opposite direction, all of the initial kinetic energy of the car is converted into potential energy stored in the spring. Therefore, we can write the equation:
Initial Kinetic Energy (KE) = Elastic Potential Energy (PE)
Using the equation for kinetic energy:
KE = 1/2 * m * v^2
Substituting the given values:
KE = 1/2 * 950 kg * (25 m/s)^2
Simplifying:
KE = 1/2 * 950 kg * 625 m^2/s^2
KE = 296,875 J
Now, equating the kinetic energy of the car to the elastic potential energy of the spring:
296,875 J = 1/2 * k * (4.0 m)^2
Simplifying:
296,875 J = 2.0 k * 16.0 m^2
Dividing both sides by 2.0 * 16.0 m^2:
296,875 J / 32.0 m^2 = k
k ≈ 9,277.34 N/m
Therefore, the spring stiffness constant of the spring is approximately 9,277.34 N/m.
To solve part (b) of the problem, we can use the equation for the time of contact between the car and the spring:
Time of contact (t) = (2 * x) / v
Where:
t is the time of contact between the car and the spring (s)
x is the compression or extension of the spring (m)
v is the initial speed of the car (m/s)
Substituting the given values:
t = (2 * 4.0 m) / 25 m/s
Simplifying:
t = 0.32 s
Therefore, the car is in contact with the spring for approximately 0.32 seconds before bouncing off in the opposite direction.