A 55 kg skateboarder accelerates down a ramp and hits the pavement moving at 5.6 m/s. If the pavement has a coefficient of friction of 0.11, how far will skateboarder glide before coming to a stop?
normal force = 55 g
friction force = .11 * 55 g
acceleration = F/m = -.11 * 55 * g /55
= -.11 g = -1.08 m/s^2
v = Vi + a t
0 = 5.6 - 1.08 t
t = 5.2 seconds to stop
go at average speed (5.6 /2) for 5.2 seconds
d = 14.5 meters
To solve this problem, we can use the concept of kinetic friction. The equation for kinetic friction is given by:
Frictional force (f) = coefficient of friction (μ) * normal force (N)
The normal force can be calculated as the product of mass (m) and acceleration due to gravity (g):
Normal force (N) = mass (m) * acceleration due to gravity (g)
The gravitational acceleration (g) is approximately 9.8 m/s^2. Therefore:
N = 55 kg * 9.8 m/s^2
N = 539 N
The frictional force can be calculated using the formula:
Frictional force (f) = μ * N
f = 0.11 * 539 N
f = 59.29 N
The frictional force is opposing the motion of the skateboarder. Using Newton's second law of motion, we know that:
Frictional force (f) = mass (m) * acceleration (a)
Therefore:
59.29 N = 55 kg * a
Solving for the acceleration (a):
a = 59.29 N / 55 kg
a ≈ 1.08 m/s^2
Using the kinematic equation:
v^2 = u^2 + 2as
where:
v = final velocity (0 m/s because the skateboarder comes to a stop)
u = initial velocity (5.6 m/s)
a = acceleration (-1.08 m/s^2)
s = distance traveled
Rearranging the equation:
s = (v^2 - u^2) / (2a)
s = (0^2 - 5.6^2) / (2 * -1.08)
s = -5.6^2 / -2.16
s ≈ 15.03 meters
Therefore, the skateboarder will glide for approximately 15.03 meters before coming to a stop.
To find the distance the skateboarder will glide before coming to a stop, we need to consider the forces acting on the skateboarder.
First, let's find the frictional force acting on the skateboarder. The frictional force can be calculated using the equation:
frictional force = coefficient of friction * normal force
The normal force can be calculated by multiplying the mass of the skateboarder by the acceleration due to gravity (9.8 m/s^2):
normal force = mass * gravity
normal force = 55 kg * 9.8 m/s^2
normal force = 539 N
frictional force = 0.11 * 539 N
frictional force = 59.29 N
Now, let's find the acceleration of the skateboarder using Newton's second law of motion:
frictional force = mass * acceleration
59.29 N = 55 kg * acceleration
acceleration = 59.29 N / 55 kg
acceleration ≈ 1.08 m/s^2
Next, we can use the kinematic equation to find the distance the skateboarder will glide before coming to a stop:
final velocity^2 = initial velocity^2 + 2 * acceleration * distance
Since the skateboarder comes to a stop, the final velocity is 0 m/s. The initial velocity is 5.6 m/s. Substituting these values into the equation:
0^2 = 5.6^2 + 2 * 1.08 * distance
Simplifying the equation:
0 = 31.36 + 2.16 * distance
Rearranging the equation to solve for distance:
2.16 * distance = -31.36
distance = -31.36 / 2.16
distance ≈ -14.5 m
The negative distance suggests that the skateboarder cannot glide in the opposite direction of the initial velocity. Therefore, we can neglect the negative sign, and the skateboarder will glide approximately 14.5 meters before coming to a stop.