A shuffleboard disk is accelerated to a speed of 5.6 m/s and released. If the coefficient of kinetic friction between the disk and the concrete court is 0.29, how far does the disk go before it comes to a stop? The courts are 15.7 m long.
To find the distance the disk goes before it comes to a stop, we need to calculate the deceleration of the disk.
The deceleration of an object can be determined using the equation:
acceleration = coefficient of kinetic friction * acceleration due to gravity
The acceleration due to gravity is approximately 9.8 m/s^2.
Substituting the given values:
acceleration = 0.29 * 9.8
acceleration = 2.842 m/s^2
Since the final velocity of the disk is 0 (because it comes to a stop), and the initial velocity is 5.6 m/s, we can use the following equation:
final velocity^2 = initial velocity^2 + 2 * acceleration * distance
Rearranging the equation to solve for distance:
distance = (final velocity^2 - initial velocity^2) / (2 * acceleration)
Plugging in the values:
distance = (0^2 - 5.6^2) / (2 * 2.842)
distance = 31.36 / 5.684
distance ≈ 5.51 m
Therefore, the disk will go approximately 5.51 meters before it comes to a stop.
To find the distance the shuffleboard disk goes before coming to a stop, we need to consider the forces acting on the disk.
First, let's determine the force of kinetic friction. The formula for the force of kinetic friction is given by:
F_k = μ_k * N
where μ_k is the coefficient of kinetic friction and N is the normal force.
Assuming the disk is on a flat surface, the normal force acting on it is equal to its weight, which can be calculated using the equation:
N = m * g
where m is the mass of the disk and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Next, we need to determine the net force acting on the disk. In this case, the only force acting on the disk is the force of kinetic friction, directed opposite to the motion of the disk.
The net force formula is given by:
F_net = m * a
where m is the mass of the disk and a is the acceleration of the disk.
Since the disk is initially accelerated to a speed of 5.6 m/s and then released, its acceleration will be in the opposite direction and can be given as:
a = -v^2 / (2d)
where v is the initial velocity (5.6 m/s) and d is the distance traveled.
By equating the force of kinetic friction to the net force, we have:
F_k = F_net
μ_k * N = m * a
Substituting the expressions for N and a, we get:
μ_k * m * g = m * (-v^2 / (2d))
Simplifying and solving for d, we have:
d = -v^2 / (2 * μ_k * g)
Substituting the given values:
d = -(5.6^2) / (2 * 0.29 * 9.8)
Calculating this expression, we find:
d ≈ -1.767 m
Since distance cannot be negative, we take the magnitude of the value:
d ≈ 1.767 m
Therefore, the shuffleboard disk travels approximately 1.767 meters before coming to a stop.
the frictional force is __ .29 g
the time to stop is __ 5.6 / (.29 g)
the average velocity is __ (5.6 + 0) / 2
the distance traveled is the average velocity divided by the time