An object is sliding along a flat surface with an initial speed of 30 m/s. What must the
coefficient of kinetic friction be between the object and the surface if the object slides to a stop in
10m? What would be the speed of the object after it had only slide half that distance?
To determine the coefficient of kinetic friction between the object and the surface, we can use the equations of motion.
1. First, we need to find the acceleration of the object. We can use the equation:
v^2 = u^2 + 2as
where:
v = final velocity (0 m/s, since the object comes to a stop)
u = initial velocity (30 m/s)
a = acceleration (unknown)
s = distance (10 m)
Rearranging the equation, we have:
a = (v^2 - u^2) / (2s)
a = (0^2 - 30^2) / (2 * 10)
a = -900 / 20
a = -45 m/s^2
2. Next, we can find the coefficient of kinetic friction (μ). The force of kinetic friction (Fk) is given by:
Fk = μ * N
where:
μ = coefficient of kinetic friction (unknown)
N = normal force
The normal force (N) is equal to the weight of the object, which can be calculated using:
N = m * g
where:
m = mass of the object
g = acceleration due to gravity (approximately 9.8 m/s^2)
Now we can find the coefficient of kinetic friction (μ). The force of kinetic friction (Fk) can also be expressed as:
Fk = m * a
Therefore,
m * a = μ * m * g
Canceling out the mass:
a = μ * g
Rearranging the equation to solve for μ:
μ = a / g
μ = -45 / 9.8
μ ≈ -4.59
The coefficient of kinetic friction between the object and the surface must be approximately -4.59.
To find the speed of the object after sliding only half the distance (5 m), we can use the equation of motion:
v^2 = u^2 + 2as
where:
v = final velocity (unknown)
u = initial velocity (30 m/s)
a = acceleration (-45 m/s^2)
s = distance (5 m)
Rearranging the equation, we have:
v^2 = 30^2 + 2(-45)(5)
v^2 = 900 + (-450)
v^2 = 450
v = √450
v ≈ 21.2 m/s
The speed of the object after sliding half the distance (5 m) would be approximately 21.2 m/s.