It's a snowy day and you're pulling a friend along a level road on a sled. You've both been taking physics, so she asks what you think the coefficient of friction between the sled and the snow is. You've been walking at a steady 1.5m/s, and the rope pulls up on the sled at a 33.0∘ angle. You estimate that the mass of the sled, with your friend on it, is 71.0kg and that you're pulling with a force of 89.0N. what answer will you give
v=const => Net force =0 =>
Fcosα=F(fr) =μN=μmg
μ= Fcosα/mg=89•cos33/71•9.8=0.1
To find the coefficient of friction between the sled and the snow, we need to consider the forces acting on the sled. In this scenario, the force you are applying to pull the sled can be broken down into two components: one parallel to the surface of the road and one perpendicular to the surface.
First, let's find the component of the force parallel to the road. We can do this by using trigonometry:
Parallel Force = Force * sin(θ)
where θ is the angle the rope makes with the ground (33.0°) and Force is the force you are applying (89.0 N).
Parallel Force = 89.0 N * sin(33.0°)
Parallel Force ≈ 47.29 N
Next, we can find the net force acting along the sled's motion. This net force is responsible for counteracting the frictional force, which can be calculated using Newton's second law:
Net Force = Mass * Acceleration
Since the sled is moving at a constant velocity, the acceleration is zero and the net force is also zero. Therefore, the frictional force must be equal in magnitude and opposite in direction to the parallel force:
Frictional Force = Parallel Force = 47.29 N
Finally, the coefficient of friction (μ) can be determined using the equation:
Frictional Force = μ * Normal Force
The normal force is equal to the weight of the sled and your friend:
Normal Force = Mass * Gravity
where Mass is the mass of the sled and your friend (71.0 kg) and Gravity is the acceleration due to gravity (9.8 m/s^2).
Normal Force = 71.0 kg * 9.8 m/s^2
Normal Force ≈ 696.8 N
Now, substituting the values into the equation, we can solve for the coefficient of friction:
47.29 N = μ * 696.8 N
μ = 47.29 N / 696.8 N
μ ≈ 0.0678
Therefore, the coefficient of friction between the sled and the snow is approximately 0.0678.