A student has tied together two sleds with a rope. Two of his friends, with masses of 55.0 kg and 60.0 kg, are riding in the sleds, one in each (55.0 kg in the back, 60.0 kg in front). Th e student uses another rope to pull the sleds with a horizontal force of magnitude 230 N. Th e sleds accelerate with a magnitude of 1.02 m/ s 2 . Th e front sled experiences a force of friction of magnitude 58.8 N. Assume that the masses of the sleds are negligible . a) What is the frictional force on the back sled? b) What is the tension in the rope connecting the sleds? c) Th e student pulling the sleds starts from rest, runs fo r 3.00 s, and then lets the sleds go. How far will the sleds travel after he lets them go?
a) The frictional force on the back sled can be found using Newton's second law. The net force acting on the back sled is the tension in the rope connecting the sleds minus the frictional force. So we have:
Net force = Tension - Frictional force
Since the sleds are accelerating, the net force is equal to the mass of the back sled times its acceleration. Rearranging the equation, we get:
Frictional force = Tension - (mass of back sled * acceleration)
Plugging in the given values, we get:
Frictional force = Tension - (55.0 kg * 1.02 m/s^2)
b) The tension in the rope connecting the sleds can be found by considering the forces acting on the front sled. The only horizontal force acting on the front sled is the tension in the rope, and this force must be equal to the product of the mass of the front sled and its acceleration. So we have:
Tension = mass of front sled * acceleration
Plugging in the given values, we get:
Tension = 60.0 kg * 1.02 m/s^2
c) To find how far the sleds will travel after the student lets them go, we need to consider their initial velocity. Since the student starts from rest and runs for 3.00 s, the velocity of the sleds when he lets them go is equal to the magnitude of the force he applied divided by the sum of the masses of his friends:
Velocity = force / (mass of back friend + mass of front friend)
Plugging in the given values, we get:
Velocity = 230 N / (55.0 kg + 60.0 kg)
Once we have the initial velocity, we can find the distance traveled using the equation:
Distance = (initial velocity * time) + (1/2 * acceleration * time^2)
Plugging in the given values, we get:
Distance = (velocity * 3.00 s) + (1/2 * 1.02 m/s^2 * (3.00 s)^2)
To find the answers to the given questions, we can use Newton's second law of motion and apply it to each sled separately. Let's break down the problem step by step.
a) Frictional force on the back sled:
We know the mass of the back sled, which is 55.0 kg. The sled experiences a force of friction, and we need to determine its magnitude. To find the force of friction, we can use Newton's second law:
Force = Mass * Acceleration
Rearranging the equation to solve for the force of friction, we get:
Force of Friction = Mass * Acceleration
Plugging in the values, we have:
Force of Friction = 55.0 kg * 1.02 m/s^2
Calculating the force of friction, we get:
Force of Friction = 56.10 N
Therefore, the frictional force on the back sled is 56.10 N.
b) Tension in the rope connecting the sleds:
To find the tension in the rope connecting the sleds, we can consider the forces acting on the front sled. We have the force of friction acting in the opposite direction and the pulling force applied by the student in the forward direction. The tension in the rope is the same as the pulling force applied by the student. Therefore, the tension in the rope is 230 N.
c) Distance traveled by the sleds after the student lets them go:
When the student lets the sleds go, there are no external forces acting on them except for the force of friction on the front sled. The sleds will continue moving with a constant velocity until the frictional force brings them to a stop. Since we know the initial velocity is zero, and the time is given as 3.00 seconds, we can calculate the distance traveled using the formula:
Distance = (1/2) * Acceleration * Time^2
Plugging in the values, we get:
Distance = (1/2) * 1.02 m/s^2 * (3.00 s)^2
Calculating the distance, we get:
Distance = 4.59 m
Therefore, the sleds will travel a distance of 4.59 meters after the student lets them go.
To solve this problem, we will use Newton's second law (F=ma) and the relationship between force, mass, and acceleration.
a) To find the frictional force on the back sled, we need to apply Newton's second law. We know that the force acting on the sleds is 230 N, the combined mass of the sleds is 55.0 kg + 60.0 kg = 115.0 kg, and the acceleration is 1.02 m/s^2. Rearranging the equation F=ma, we can solve for the frictional force on the back sled:
Frictional force = Force - mass * acceleration
Frictional force = 230 N - 115.0 kg * 1.02 m/s^2
Frictional force = 230 N - 117.3 N
Frictional force = 112.7 N
The frictional force on the back sled is 112.7 N.
b) To find the tension in the rope connecting the sleds, we need to consider the forces acting on the sleds. The tension in the rope connecting the sleds is equal to the force required to accelerate the sleds. Since both sleds are accelerating together, the total mass is 115.0 kg, and the acceleration is 1.02 m/s^2, we can use Newton's second law to calculate the tension:
Tension = mass * acceleration
Tension = 115.0 kg * 1.02 m/s^2
Tension = 117.3 N
The tension in the rope connecting the sleds is 117.3 N.
c) To find how far the sleds will travel after the student lets them go, we need to consider the initial velocity, the time, and the constant acceleration. The initial velocity is 0 m/s since the student starts from rest. The time is given as 3.00 s, and the acceleration is 1.02 m/s^2. We can use the following kinematic equation to solve for the distance traveled:
Distance = Initial velocity * time + 0.5 * acceleration * time^2
Distance = 0 * 3.00 s + 0.5 * 1.02 m/s^2 * (3.00 s)^2
Distance = 4.59 m
The sleds will travel a distance of 4.59 meters after the student lets them go.