during a hockey game a puck of mass .3 kg is given an inital speed of 12m/s. it slides 50 m o the ice before stopping. Assume the puck decelerates at a constant rate. what is the coefficient of friction betweent the puck and the ice?
I know that it would be a very small answer since there isn't a lot of friction between ice and an object.
I know that the normal force is equal to F_g which means it equals 3N (.3kg*10m/s^2).
But now i have two unknowns I don't know mu and i don't know F_f.
F_f=mu(F_n)
What am i doing wrong?
I would do it with energy
1/2 m v%2= frictionforce*distance
and friction force = mu*mg
solve for mu. Notice mass divide out.
To solve for the coefficient of friction between the puck and the ice, you need to use the given information and make use of Newton's second law of motion. Here's how you can approach the problem:
1. Identify the forces acting on the puck: In this case, the two main forces are the force of gravity (Fg) and the force of friction (Ff).
2. Determine the net force acting on the puck: Since the puck is decelerating (slowing down), the net force is in the opposite direction of its initial velocity. Therefore, the net force can be calculated using Newton's second law, Fnet = ma, where Fnet is the net force, m is the mass of the puck, and a is the acceleration.
Given: Mass (m) = 0.3 kg, Initial velocity (v) = 12 m/s, Distance (d) = 50 m
Final Velocity (vf) = 0 m/s (since it stops)
Using the kinematic equation vf^2 = vi^2 + 2ad (where vi is the initial velocity and a is the acceleration), you can find the acceleration:
0^2 = 12^2 + 2a(50)
0 = 144 + 100a
-144 = 100a
a = -1.44 m/s^2 (negative because it's decelerating)
3. Calculate the net force: Fnet = ma
Fnet = 0.3 kg × (-1.44 m/s^2)
Fnet = -0.432 N
4. Determine the frictional force: The frictional force can be calculated using the equation Ff = μFn, where μ is the coefficient of friction and Fn is the normal force (which is equal to the weight of the object since it is on a horizontal surface).
Fn = Fg = mg (where g is the acceleration due to gravity, approximately 10 m/s^2)
Fn = 0.3 kg × 10 m/s^2
Fn = 3 N
Ff = μFn
-0.432 N = μ × 3 N
μ = -0.432 N / 3 N
μ ≈ -0.144
The negative value of the coefficient of friction indicates that the force of friction is acting in the opposite direction of motion. However, coefficients of friction are typically positive values, representing the ratio of the force of friction to the normal force. Therefore, in this case, the coefficient of friction between the puck and the ice would be approximately 0.144.
Note: It is unusual to have a negative coefficient of friction. Typically, the negative sign indicates that there might be some other force involved that is opposing the motion of the puck. However, in this case, with the given information and assumption of a constant deceleration, the negative coefficient of friction is the result.