A 110g hockey puck sent sliding over ice is stopped by a frictional force on it from the ice a) if its intial speed is 6.0m/s, what is the magnitude of the frictional force? b) what is the coefficient of friction between the puck and the ice?

I am unsure how to solve the first part of the problem because they give met only intial velocity and distance but no time. I assuming that 15 m stands for meters. Please give me some direction thank you

If you know the distance, you can use:

Vf^2 = vo^2 + 2ad,

Vf is final velocity; vo is initial velocity, and d is distance. Once you know a, you can calculate the force.

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To solve part a) of the problem, you can use the equation Vf^2 = vo^2 + 2ad, where Vf is the final velocity, vo is the initial velocity, d is the distance, and a is the acceleration.

Given:
- Mass of the hockey puck (m) = 110g = 0.11 kg
- Initial velocity (vo) = 6.0 m/s
- Distance (d) = unknown

Since the hockey puck is being stopped by the frictional force, the final velocity (Vf) will be 0 m/s.

Substituting the values into the equation, we get:
0^2 = 6.0^2 + 2a(d)

Simplifying the equation, we have:
0 = 36 + 2ad

Since the puck is being stopped, we know the final velocity is 0, so the equation becomes:
0 = 6.0^2 + 2ad

Now, we need to find the distance (d) for the given initial velocity. We can rearrange the equation to solve for d:

d = (0 - 6.0^2) / (2a)
d = -36 / (2a)
d = -18 / a

Since we don't have the value for acceleration (a), we cannot calculate the distance (d) at this point. We need additional information or equations to proceed.

Regarding your assumption, yes, 15 m stands for 15 meters.

For part b) of the problem, to determine the coefficient of friction (μ) between the puck and the ice, we need to know the magnitude of the frictional force (F_friction) and the normal force (F_normal) acting on the puck.

The equation to calculate the magnitude of the frictional force is:
F_friction = μ * F_normal

However, without knowing either the value of the frictional force or the normal force, we cannot calculate the coefficient of friction (μ) at this point.

To proceed with solving part b), we would need additional information such as the surface area in contact between the puck and the ice or the angle at which the puck is sliding, which could help determine the normal force or the frictional force.

To solve part a) of the problem, you can use the equation Vf^2 = Vo^2 + 2ad. Although the time is not given, you can still find the solution by using the distance instead.

Given:
Vo = 6.0 m/s (initial velocity of the puck)
d = 15 m (distance traveled by the puck)

First, let's calculate the final velocity (Vf). Rearranging the equation:

Vf^2 = Vo^2 + 2ad

Vf^2 = (6.0 m/s)^2 + 2 * a * 15 m

Since we are looking for the magnitude of the frictional force, we know that the final velocity will be zero when the puck comes to a stop. Therefore, Vf = 0.

0 = (6.0 m/s)^2 + 2 * a * 15 m

Now, solve for the acceleration (a):

(6.0 m/s)^2 = 2 * a * 15 m

36 m^2/s^2 = 30 a

a = (36 m^2/s^2) / 30

a ≈ 1.2 m/s^2

Now that you know the acceleration, you can calculate the magnitude of the frictional force using Newton's second law, which states that F = ma, where F is the force, m is the mass, and a is the acceleration.

Given:
m = 110 g = 0.11 kg (mass of the puck)
a = 1.2 m/s^2 (acceleration)

F = m * a
F = 0.11 kg * 1.2 m/s^2

Therefore, the magnitude of the frictional force is approximately F ≈ 0.132 N.

As for part b), to calculate the coefficient of friction between the puck and the ice, you need to use the equation:

F = μ * N

Where F is the frictional force, μ is the coefficient of friction, and N is the normal force.

In this case, the normal force is equal to the weight of the puck, which can be calculated as:

N = m * g

Given:
m = 110 g = 0.11 kg (mass again, in SI units)
g ≈ 9.8 m/s^2 (acceleration due to gravity)

N = 0.11 kg * 9.8 m/s^2

Now substitute the known values into the equation for frictional force:

0.132 N = μ * (0.11 kg * 9.8 m/s^2)

Solve for the coefficient of friction (μ):

μ = (0.132 N) / (0.11 kg * 9.8 m/s^2)

Therefore, the coefficient of friction between the puck and the ice is approximately μ ≈ 0.122.