A 1.80 kg block slides on a frictionless horizontal surface. The block hits a spring with a speed of 2.00 m/s and compresses it a distance of 11.0 cm before coming to rest. What is the force constant of the spring?

When compressed the maximum amount , the potential eneegy stored in the spring equals the initial kinetic energy. Thus

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

k = M(Vo/X)^2

Vo = 0.11 m

Solve for k

k will have units of N/m

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

Hooke's Law can be expressed as:
F = -kx

Where F is the force exerted by the spring, k is the force constant (also known as the spring constant), and x is the displacement from the equilibrium position.

In this problem, we are given the mass of the block (1.80 kg), the initial speed of the block (2.00 m/s), and the distance the spring is compressed (11.0 cm).

We can start by calculating the initial kinetic energy of the block, since all of this energy is converted into potential energy stored in the compressed spring.

The initial kinetic energy (KE) can be calculated using:
KE = 0.5 * mass * velocity^2

Plugging in the values, we get:
KE = 0.5 * 1.80 kg * (2.00 m/s)^2

Simplifying, we find:
KE = 3.60 J

This initial kinetic energy is converted into potential energy stored in the spring, which can be calculated using:
PE = 0.5 * k * x^2

Plugging in the given distance of compression (11.0 cm = 0.11 m) and the initial kinetic energy (3.60 J), we get:
3.60 J = 0.5 * k * (0.11 m)^2

Simplifying further, we have:
3.60 J = 0.5 * k * 0.0121 m^2

Now, we can solve for k by isolating it in the equation:
k = (2 * PE) / x^2

Plugging in the values, we get:
k = (2 * 3.60 J) / 0.0121 m^2

Evaluating this expression, we find:
k = 593.39 N/m

Therefore, the force constant (spring constant) of the spring is approximately 593.39 N/m.