On a frozen pond, a 7 kg sled is given a kick
that imparts to it an initial speed of 3.1 m/s.
The coefficient of kinetic friction between sled
and ice is 0.15.
The acceleration of gravity is 9.8 m/s2 .
Find the distance the sled moves before
coming to rest.
Answer in units of m.
Ke = (1/2) m v^2 = m (4.8) Joules
F = .15 m g = m ( 1.47) Newtons
F d = Ke
d = m (4.8)/[ m(1.47)]
d = 4.8/1.47
3.2687 meters
To find the distance the sled moves before coming to rest, we will use the concept of kinetic friction and the equations of motion.
Step 1: Determine the force of kinetic friction acting on the sled.
The force of kinetic friction can be calculated using the formula:
Fk = μk * N
where Fk is the force of kinetic friction, μk is the coefficient of kinetic friction, and N is the normal force.
In this case, the normal force is equal to the weight of the sled, which can be calculated as:
N = m * g
where m is the mass of the sled and g is the acceleration due to gravity.
Substituting the given values:
m = 7 kg
g = 9.8 m/s^2
N = 7 kg * 9.8 m/s^2 = 68.6 N
μk = 0.15
Fk = 0.15 * 68.6 N = 10.29 N
Step 2: Calculate the deceleration of the sled.
The force of kinetic friction causes the sled to decelerate. The deceleration can be calculated using Newton's second law:
F = m * a
where F is the net force acting on the sled, m is the mass of the sled, and a is the deceleration.
In this case, the net force is equal to the force of kinetic friction acting in the opposite direction.
So we have:
F = Fk = 10.29 N
m = 7 kg
10.29 N = 7 kg * a
Solving for a, we get:
a = 10.29 N / 7 kg = 1.47 m/s^2
Step 3: Use the equation of motion to calculate the distance traveled by the sled.
The equation of motion that relates distance (d), initial velocity (v0), final velocity (vf), and acceleration (a) is:
vf^2 = v0^2 + 2ad
In this case, the final velocity is 0 m/s (since the sled comes to rest), the initial velocity is 3.1 m/s, and the acceleration is -1.47 m/s^2 (negative because it acts in the opposite direction to the initial velocity).
Plugging in these values, we can solve for the distance traveled (d):
0^2 = (3.1 m/s)^2 + 2 * -1.47 m/s^2 * d
9.61 m^2/s^2 = 2.94 m^2/s^2 - 2.94 m^2/s^2 * d
9.61 m^2/s^2 - 2.94 m^2/s^2 = -2.94 m^2/s^2 * d
6.67 m^2/s^2 = 2.94 m^2/s^2 * d
d = 6.67 m^2/s^2 / 2.94 m^2/s^2
d = 2.27 m
Therefore, the sled moves a distance of 2.27 meters before coming to rest.
To find the distance the sled moves before coming to rest, we can use the concepts of Newton's Laws of Motion and the equations of motion.
First, let's analyze the forces acting on the sled:
1. The force of gravity (Fg) acts vertically downward with a magnitude of mg, where m is the mass of the sled and g is the acceleration due to gravity.
Fg = m * g
2. The force of friction (Ff) opposes the motion of the sled and depends on the coefficient of kinetic friction (μk) and the normal force (Fn). The normal force is equal to the weight of the sled since it is on a horizontal surface.
Fn = Fg = m * g
Ff = μk * Fn
Now, let's calculate the force of friction:
Ff = μk * Fn
= μk * m * g
Next, we can use Newton's second law to relate the net force to the acceleration:
Net force (Fnet) = m * a
For the sled, in the direction of motion:
Fnet = Fk - Ff
Since the sled comes to rest, the net force is 0:
0 = Fk - Ff
Substituting the expressions for Fk and Ff:
0 = m * a - μk * m * g
Now, solve for the acceleration (a):
a = μk * g
Substitute the given values:
a = 0.15 * 9.8 m/s^2
a = 1.47 m/s^2
Next, we can use the kinematic equation to determine the distance (d) the sled moves before coming to rest:
v^2 = u^2 + 2ad
Where:
v = final velocity (0 m/s since the sled comes to rest)
u = initial velocity (3.1 m/s)
a = acceleration (1.47 m/s^2)
d = distance
Rearranging the equation:
0 = (3.1 m/s)^2 + 2 * 1.47 m/s^2 * d
Simplify the equation:
0 = 9.61 m^2/s^2 + 2.94 m^2/s^2 * d
Solving for d:
2.94 m^2/s^2 * d = -9.61 m^2/s^2
d = (-9.61 m^2/s^2) / (2.94 m^2/s^2)
d = -3.27 m
Since distance can't be negative, we take the magnitude:
d = 3.27 m
Therefore, the sled moves approximately 3.27 meters before coming to rest.