A hockey player hits a puck with his stick, giving the puck an initial speed of 5.0 m/s. If the puck slows uniformly and comes to rest in a distance of 20 m, what is the coefficient of kinetic friction between the ice and the puck?

idk bruh

i have no idea

Frictionforce*distance=1/2 m vi^2

mu*mg*distance=1/2 m vi^2
solve for mu.

.05

To determine the coefficient of kinetic friction between the ice and the puck, you can use the principles of mechanical energy.

First, let's identify the various quantities involved in the problem:

- Initial speed of the puck (u) = 5.0 m/s
- Final speed of the puck (v) = 0 m/s (since it comes to rest)
- Distance traveled by the puck (s) = 20 m

Now, we can use the equation for uniformly decelerating motion:

v² = u² - 2as

where:
- u is the initial speed (5.0 m/s),
- v is the final speed (0 m/s),
- a is the acceleration (negative, since the puck is slowing down),
- s is the distance traveled (20 m).

Rearranging the equation, we get:

a = (v² - u²) / (2s)

Substituting the values into the equation:

a = (0² - 5.0²) / (2 * 20)

Calculating:

a = (-25.0) / 40
a = -0.625 m/s²

Since the acceleration is negative, it means there is an opposing force acting in the direction of motion. In this case, it is due to the kinetic friction between the puck and the ice.

We can use the equation for kinetic friction to find the coefficient of kinetic friction (μ):

Fk = μ * N

where:
- Fk is the force of kinetic friction,
- μ is the coefficient of kinetic friction, which we need to find,
- N is the normal force, which is the downward force exerted by the puck's weight.

The normal force is equal to the puck's weight when it is on a horizontal surface, so:

N = mg

where:
- m is the mass of the puck,
- g is the acceleration due to gravity (approximately 9.8 m/s²).

However, we can eliminate the need to calculate the mass by dividing both sides of the equation by m, giving:

a = g

Now, we can substitute the value of acceleration (a) into the equation:

Fk = μ * g

Substituting our previously calculated acceleration:

Fk = μ * (-0.625)

Since the normal force (N) cancels out from both sides, we don't need its value.

The force of kinetic friction is given by Newton's second law:

Fk = ma

Substituting the value of acceleration:

μ * (-0.625) = (-0.625)

Finally, solving for the coefficient of kinetic friction (μ):

μ = (-0.625) / (-0.625)

The negative sign cancels out, resulting in:

μ = 1.0

Therefore, the coefficient of kinetic friction between the ice and the puck is 1.0.