A 880 kg car strikes a huge spring at a speed of 20 m/s (Fig. 11-50), compressing it 15.0 m.

(a) What is the spring constant of the spring? (Found this, it was 1564.44)

(b) How long is the car in contact with the spring before it bounces off in the opposite direction? <--Need help with this one, thanks!

f=ma => kx/m=v/t slove for t

200

To solve this problem, we need to use the principle of conservation of mechanical energy. When the car hits the spring, it transfers its kinetic energy to potential energy stored in the spring, which is then used to propel the car back in the opposite direction.

Let's start by finding the potential energy stored in the spring when it is compressed by 15.0 m. The formula for potential energy stored in a spring is given by:

Potential energy (U) = (1/2) * k * x^2

where k is the spring constant and x is the displacement of the spring. We can rearrange the formula to solve for the spring constant:

k = (2 * U) / x^2

In this case, the potential energy (U) can be determined using the initial kinetic energy of the car. The formula for kinetic energy is given by:

Kinetic energy (K) = (1/2) * m * v^2

where m is the mass of the car and v is its velocity. Rearranging the formula, we can solve for U:

U = K = (1/2) * m * v^2

Plugging in the values:

m = 880 kg
v = 20 m/s

U = K = (1/2) * 880 kg * (20 m/s)^2 = 176,000 J

Now, substituting the values into the first formula, we can find the spring constant:

k = (2 * 176,000 J) / (15.0 m)^2 = 1564.44 N/m (rounded to two decimal places)

For part (b) - How long is the car in contact with the spring before it bounces off in the opposite direction?

To find the time of contact, we can use the equation of motion for a spring:

x = (1/2) * a * t^2

Where x is the displacement, a is the acceleration, and t is the time. For a spring, the acceleration is given by:

a = F / m

where F is the force exerted by the spring and m is the mass of the car.

From Hooke's Law, we know that the force exerted by the spring is related to the displacement:

F = k * x

Substituting the values:

k = 1564.44 N/m
x = 15.0 m
m = 880 kg

F = k * x = 1564.44 N/m * 15.0 m = 23,466.6 N

Now we can calculate the acceleration:

a = F / m = 23,466.6 N / 880 kg = 26.68 m/s^2 (rounded to two decimal places)

Finally, we can find the time of contact using the equation of motion:

15.0 m = (1/2) * 26.68 m/s^2 * t^2

Simplifying and solving for t:

t^2 = (2 * 15.0 m) / 26.68 m/s^2
t^2 = 0.89588 s^2

Taking the square root:

t = √(0.89588 s^2) = 0.947 s (rounded to three decimal places)

Therefore, the car is in contact with the spring for approximately 0.947 seconds before bouncing off in the opposite direction.