Suppose that the coefficient of friction between your feet and the floor, while wearing socks, is 0.250. Knowing this, you decide to get a running start and then slide across the floor. If your speed is 3.00 m/s when you start to slide, what distance d will you slide before stopping?
To find the distance you will slide before stopping, we can use the principles of physics and the equations of motion. The key equation we will use is the equation for the distance traveled during uniform deceleration:
d = (v^2 - u^2) / (2a)
Where:
d = distance traveled
v = final velocity (0 m/s, in this case, since you stop)
u = initial velocity (3.00 m/s, in this case)
a = acceleration (opposite in direction to the motion)
From the given information, we know the coefficient of friction between your feet and the floor is 0.250. The frictional force can be calculated using the equation:
f = μ * N
Where:
f = force of friction
μ = coefficient of friction
N = Normal force
The normal force, N, is equal to the weight of the person, which can be calculated using:
N = m * g
Where:
m = mass of the person
g = acceleration due to gravity (approximately 9.8 m/s^2)
We also know that the force of friction can be calculated using Newton's second law:
f = m * a
Since the mass of the person cancels out in the equation, we can equate the two expressions for the force of friction:
μ * N = m * a
Substituting the expression for N, we get:
μ * m * g = m * a
Simplifying, the mass cancels out, leaving:
μ * g = a
Now we have the acceleration, a, which is equal to the acceleration due to gravity multiplied by the coefficient of friction:
a = μ * g
Substituting this value of a into the equation for distance, we get:
d = (v^2 - u^2) / (2 * (μ * g))
Plugging in the given values, we have:
d = (0^2 - 3.00^2) / (2 * (0.250 * 9.8))
Simplifying and calculating the expression, we find:
d = -9.18 m/s^2
Since distance cannot be negative, we take the absolute value of the result:
d ≈ 37.24 meters
Therefore, you will slide approximately 37.24 meters before coming to a stop.