A 17-kg sled is being pulled along the horizontal snow-covered ground by a horizontal force of 29 N. Starting from rest, the sled attains a speed of 2.1 m/s in 9.6 m. Find the coefficient of kinetic friction between the runners of the sled and the snow.
V^2 = Vo^2 + 2a*d
a=(V^2-Vo^2)/2
a = ((2.1)^2-0)/19.2=0.230 m/s^2
Fn = m*a
(Fap-Fk) = m*a
(29-Fk) = 17*0.230
29-Fk = 3.91
Fk = 29-3.91 = 25.1 N.
Ws = m*g = 17kg * 9.8N/kg = 167 N. =
Wt. of sled.
u = Fk/Ws = 25.1/167 = 0.150 = Coefficient of kinetic friction.
Well, well, well, looks like we've got ourselves a slippery situation here, quite literally! So, we have a sled being pulled on snow-covered ground. It starts from rest and attains a speed of 2.1 m/s in a distance of 9.6 m. The pulling force is 29 N. We need to find the coefficient of kinetic friction between the sled and the snow.
To figure this out, we'll need to do a little calculation dance. The net force acting on the sled can be found by using Newton's second law, which tells us that force is equal to mass times acceleration. In this case, our acceleration is the change in speed divided by the change in distance.
So, acceleration equals (final velocity - initial velocity) divided by distance. Plug in the numbers: (2.1 m/s - 0 m/s) / 9.6 m = 0.21875 m/s².
Now, let's calculate the net force acting on the sled. We know that the pulling force is 29 N and the force of friction is in the opposite direction. Therefore, the net force is the pulling force minus the force of friction. And this net force is equal to mass times acceleration.
Substituting the values, we get 29 N - (17 kg)(0.21875 m/s²) = (17 kg)(acceleration). Simplifying the equation gives us 29 N - 3.7125 N = (17 kg)(acceleration), which leads us to 25.2875 N = (17 kg)(acceleration).
Now, let's find the force of friction. We know that the force of friction is equal to the coefficient of friction multiplied by the normal force. In this case, the normal force is equal to the weight of the sled (mass times gravity).
The weight of the sled is (17 kg)(9.8 m/s²) = 166.6 N. So, the force of friction is equal to the coefficient of friction times 166.6 N.
Finally, we can solve for the coefficient of kinetic friction by setting the force of friction equal to the net force we calculated earlier.
166.6 N times the coefficient of kinetic friction = 25.2875 N.
Therefore, the coefficient of kinetic friction is 25.2875 N / 166.6 N, which is approximately 0.152. Ta-da!
To find the coefficient of kinetic friction between the sled and the snow, we can use the formula:
\(F_{friction} = \mu_k \cdot F_{normal}\)
Where:
\(F_{friction}\) is the force of friction,
\(\mu_k\) is the coefficient of kinetic friction, and
\(F_{normal}\) is the normal force.
Step 1: Calculate the normal force.
The normal force is the force exerted by a surface that is perpendicular to it. In this case, it is equal to the weight of the sled, since there is no vertical acceleration.
\(F_{normal} = m \cdot g\)
Where:
\(m\) is the mass of the sled, and
\(g\) is the acceleration due to gravity (9.8 m/s^2).
Given that the mass of the sled is 17 kg:
\(F_{normal} = 17 kg \cdot 9.8 m/s^2\)
\(F_{normal} = 166.6 N\) (rounded to one decimal place).
Step 2: Calculate the force of friction.
The force of friction can be calculated using Newton's second law:
\(F_{friction} = m \cdot a\)
Where:
\(m\) is the mass of the sled, and
\(a\) is the acceleration of the sled.
Given that the initial velocity of the sled is 0 m/s, the final velocity is 2.1 m/s, and the distance traveled is 9.6 m, we can use the following kinematic equation to find the acceleration:
\(v^2 = u^2 + 2as\)
Where:
\(v\) is the final velocity,
\(u\) is the initial velocity,
\(a\) is the acceleration, and
\(s\) is the distance traveled.
Plugging in the values:
\(2.1^2 = 0^2 + 2a \cdot 9.6\)
\(4.41 = 19.2a\)
\(a = \frac{4.41}{19.2}\)
\(a = 0.229 m/s^2\) (rounded to three decimal places).
Now, we can calculate the force of friction:
\(F_{friction} = m \cdot a\)
\(F_{friction} = 17 kg \cdot 0.229 m/s^2\)
\(F_{friction} = 3.883 N\) (rounded to three decimal places).
Step 3: Calculate the coefficient of kinetic friction.
Finally, we can substitute the known values into the formula for friction:
\(F_{friction} = \mu_k \cdot F_{normal}\)
\(3.883 N = \mu_k \cdot 166.6 N\)
To solve for \(\mu_k\), divide both sides of the equation by \(166.6 N\):
\(\mu_k = \frac{3.883 N}{166.6 N}\)
\(\mu_k \approx 0.023\) (rounded to three decimal places).
Therefore, the coefficient of kinetic friction between the runners of the sled and the snow is approximately 0.023.
To find the coefficient of kinetic friction between the runners of the sled and the snow, we can start by analyzing the forces acting on the sled.
First, we can find the net force acting on the sled. The net force is equal to the product of the mass of the sled and its acceleration.
Given:
Mass of the sled (m) = 17 kg
Acceleration of the sled (a) = ?
Force applied (F) = 29 N
According to Newton's second law of motion, F = m * a. Rearranging the equation, we can solve for acceleration: a = F / m.
Plugging in the known values, we get:
a = 29 N / 17 kg
a ≈ 1.71 m/s²
Next, we need to determine the coefficient of kinetic friction (μ) between the sled and the snow. The kinetic friction force (Fk) can be calculated using the formula Fk = μ * N, where N is the normal force.
The normal force (N) is the force exerted perpendicular to the contact surface. In this case, since the sled is pulled horizontally, the normal force is equal to the weight of the sled, which can be found using the formula N = m * g, where g is the acceleration due to gravity (approximately 9.8 m/s²).
Plugging in the given values, we have:
N = 17 kg * 9.8 m/s²
N ≈ 166.6 N
Now, we can calculate the kinetic friction force:
Fk = μ * N
Using the known value of the applied force (29 N), we can write the equation of motion:
F - Fk = m * a
Substituting the values we have:
29 N - μ * 166.6 N = 17 kg * 1.71 m/s²
Simplifying the equation and solving for μ, we get:
μ ≈ (29 N - 17 kg * 1.71 m/s²) / 166.6 N
Calculating this, we find:
μ ≈ 0.048
Therefore, the coefficient of kinetic friction between the runners of the sled and the snow is approximately 0.048.