A polar bear cub slides (starting from rest) 31 m toward his mother down a snow-covered 37^\circ slope starting from rest. At the end of the 31 m, the cub’s speed is 4.7 m/s. Find the coefficient of friction between the cub and the snow-covered surface.
How far down did he go
31 sin 37 = 31 * 0.602 = 18.7 meters down
so
loss of potential energy = m g * 18.7
gain in kinetic energy = (1/2) m v^2 = m * 4.7^2 / 2 = 11 m
so energy wasted on friction = m (18.7 g - 11 )
that is the work done by the friction force
mu m g cos 37 * 31 = m (18.7 g - 11)
say g = 9.81
mu * 9.81 * 31 * cos 37 = 18.7 * 9.81 - 11
243 mu = 172
mu = 0.71
To find the coefficient of friction between the cub and the snow-covered surface, we can start by analyzing the forces acting on the cub.
The force of gravity can be divided into two components: one parallel to the slope and one perpendicular to the slope. The component parallel to the slope is responsible for the acceleration of the cub down the slope, while the perpendicular component is balanced by the normal force.
Let's call the parallel component of the force of gravity F_parallel. Since the only other horizontal force acting on the cub is friction, we can equate F_parallel to the force of friction.
F_parallel = force of friction ...(1)
Now, let's find the value of F_parallel. We know that the cub's speed is 4.7 m/s at the end of the 31 m slide. We also know that the cub started from rest, so we can find the acceleration using the kinematic equation:
v^2 = u^2 + 2as
Here, v is the final velocity (4.7 m/s), u is the initial velocity (0 m/s), a is the acceleration, and s is the displacement (31 m downhill).
Plugging in the values:
(4.7 m/s)^2 = (0 m/s)^2 + 2 * a * 31 m
Simplifying:
22.09 m^2/s^2 = 62 a
Dividing by 62:
a = 0.356 m/s^2
Now that we have the acceleration, we can find F_parallel using Newton's second law:
F_parallel = m * a
Here, m is the mass of the cub. Let's assume the mass of the cub is m_cub.
Substituting the mass and acceleration:
F_parallel = m_cub * 0.356 m/s^2 ...(2)
Next, let's consider the normal force. The normal force is equal in magnitude but opposite in direction to the perpendicular component of the force of gravity. Since the slope is inclined at an angle of 37 degrees, the perpendicular component of the force of gravity is given by:
F_perpendicular = m_cub * g * cos(37 degrees)
Here, g is the acceleration due to gravity (9.8 m/s^2).
Now, since the perpendicular component of the force of gravity is balanced by the normal force, we can say:
F_perpendicular = normal force
We also know that the normal force is equal in magnitude but opposite in direction to the force of friction, so we can rewrite equation (1) as:
F_parallel = F_perpendicular * μ ...(3)
where μ is the coefficient of friction.
Now, substituting the expressions for F_parallel and F_perpendicular from equations (2) and (3) respectively:
m_cub * 0.356 m/s^2 = m_cub * g * cos(37 degrees) * μ
Simplifying:
0.356 m/s^2 = g * cos(37 degrees) * μ
Substituting the value of g and simplifying:
0.356 m/s^2 = (9.8 m/s^2) * cos(37 degrees) * μ
Finally, solving for μ:
μ = 0.356 m/s^2 / (9.8 m/s^2 * cos(37 degrees))
Using a calculator:
μ ≈ 0.079
Therefore, the coefficient of friction between the cub and the snow-covered surface is approximately 0.079.