A 1350 kg car rolling on a horizontal surface has a speed of 40 km/hr when it strikes a horizontal coiled spring and is brought to rest in a distance of 5.5 m. What is the spring constant (in N/m) of the spring?

Initial kinetic energy = compressed spring potential energy.

(1/2)M*Vo^2 = (1/2)k*X^2

Convert Vo to m/s and solve for k.
X = 5.5m is the max. spring deflection.

5509

To find the spring constant of the coiled spring, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position.

The equation for Hooke's Law is F = k * x, where F is the force, k is the spring constant, and x is the displacement.

First, let's convert the speed of the car from km/hr to m/s:
40 km/hr = (40 * 1000 m) / (60 * 60 s) = 11.11 m/s

Next, we need to calculate the deceleration of the car. Since the car comes to rest, its final velocity is 0 m/s. We can use the kinematic equation:
v^2 = u^2 + 2as
where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement.

Rearranging the equation, we find:
a = (v^2 - u^2) / (2s)
= (0 - 11.11^2) / (2 * 5.5)
= -137.12 m/s^2

The negative sign indicates that the acceleration is in the opposite direction of the initial velocity, causing deceleration.

Now, we can calculate the force exerted by the spring using Newton's second law:
F = m * a
= 1350 kg * (-137.12 m/s^2)
= -185,112 N

Since the spring force is in the opposite direction of the car's motion, it is negative.

Finally, we can determine the spring constant by rearranging Hooke's Law equation:
k = F / x
= -185,112 N / 5.5 m
= -33,838.2 N/m

The spring constant of the coiled spring is approximately -33,838.2 N/m.