1.) If the curling stone was initially sliding at 2.97 m/s and stopped precisely 10.8 meters from its launch point, then the coefficient of friction (the ratio between the magnitude of the frictional force and the magnitude of the normal force) was?
3.) Two curling contestants dispute possession of a stone of mass 15.0 kg. The mass of contestant A is 56.2 kg, whereas that of contestant B is 105.2 kg. Contestant A pulls on the stone with a horizontal force of 19.5 N, but is unable to break contestant B’s grip. The result is a stalemate (the stone does not accelerate).
a) Neglecting (for now) frictional forces between the stone and the ice, what is the horizontal force that contestant B exerts on the stone? [Note: Indicate a pull (a force in contestant B’s direction) by a positive sign and a push (a force in contestant A’s direction) by a negative sign.]
=?N
b) What is the minimum coefficient of static friction required between the contestants and the ice (assume that it is the same for both contestants)?
To find the coefficient of friction, we need to use the equation of motion for the first question and analyze the forces in the second question.
1.) To find the coefficient of friction, we can use the equation of motion:
v^2 = u^2 + 2as
Where,
v = final velocity (0 m/s since the stone stops)
u = initial velocity (2.97 m/s)
a = acceleration due to friction
s = displacement (10.8 m)
Rearranging the equation:
a = (v^2 - u^2) / (2s)
a = (0^2 - 2.97^2) / (2 * 10.8)
a = -2.97^2 / (2 * 10.8)
a = -0.82286 m/s^2
The negative sign indicates that the frictional force acts in the opposite direction to the sliding direction.
Now, we can calculate the magnitude of the frictional force using:
frictional force = mass * acceleration
Assuming the mass of the stone is 1 kg (just for calculations), the frictional force can be calculated as:
frictional force = 1 kg * -0.82286 m/s^2
frictional force = -0.82286 N
To find the magnitude of the normal force, we can assume that the normal force is equal to the weight of the stone. Let's say the stone weighs 10 N (again, for calculations).
The coefficient of friction is defined as the ratio between the magnitude of the frictional force and the magnitude of the normal force:
coefficient of friction = |frictional force| / |normal force|
coefficient of friction = |-0.82286 N| / |10 N|
coefficient of friction = 0.082286
So, the coefficient of friction is approximately 0.082286.
3.) Now let's move on to the second question.
a) To find the horizontal force that contestant B exerts on the stone, we can apply Newton's third law of motion. According to Newton's third law, the force exerted by contestant A on the stone and the force exerted by contestant B on the stone are equal in magnitude but opposite in direction.
Since contestant A pulls on the stone with a horizontal force of 19.5 N, contestant B exerts an equal but opposite force on the stone. Thus, contestant B exerts a force of -19.5 N (negative sign indicates a push force in contestant A's direction).
b) To find the minimum coefficient of static friction required between the contestants and the ice, we need to consider the fact that there is no acceleration of the stone. This means that the force exerted by contestant B (frictional force) must be equal to the force exerted by contestant A (19.5 N).
Using the relation between static friction and normal force (fs ≤ μs * fn), where fs is the frictional force, μs is the coefficient of static friction, and fn is the normal force:
19.5 N = μs * fn
Since the normal force on the stone is equal to its weight (15 kg * 9.8 m/s^2 = 147 N), we can rearrange the equation:
μs = 19.5 N / 147 N
μs ≈ 0.13265
So, the minimum coefficient of static friction required between the contestants and the ice is approximately 0.13265.