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). The student uses another rope to pull the sleds with a horizontal force of magnitude 230 N. The sleds accelerate with a magnitude of 1.02 m/ s^2. The 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) The student pulling the sleds starts from rest, runs for 3.00 s, and then lets the sleds go. How far will the sleds travel aft er he lets them go?
a) To find the frictional force on the back sled, we need to first find the net force acting on it. The net force can be determined using Newton's second law:
Net force = mass * acceleration
The mass of the back sled is 55.0 kg and the acceleration is 1.02 m/s^2:
Net force = 55.0 kg * 1.02 m/s^2 = 56.1 N
Since the sled is being pulled to the right with a force of 230 N and experiencing friction to the left, the frictional force on the back sled can be found by subtracting the force of tension from the net force:
Frictional force on back sled = Net force - Force of tension
We don't know the force of tension yet, so let's solve for it:
Force of tension = Net force - Frictional force on back sled
Force of tension = 230 N - 56.1 N = 173.9 N
Now we can substitute this value back into the equation to find the frictional force on the back sled:
Frictional force on back sled = Net force - Force of tension
Frictional force on back sled = 56.1 N - 173.9 N = -117.8 N
Therefore, the frictional force on the back sled is -117.8 N (opposite to the direction of motion).
b) The tension in the rope connecting the sleds can be determined by considering the equilibrium of forces on the front sled. The equation for the front sled is:
Net force on front sled = Force of tension - Force of friction
We know the net force on the front sled is given by:
Net force on front sled = mass * acceleration
Net force on front sled = 60.0 kg * 1.02 m/s^2 = 61.2 N
By rearranging the equation, we can solve for the force of tension:
Force of tension = Net force on front sled + Force of friction
Force of tension = 61.2 N + 58.8 N = 120.0 N
Therefore, the tension in the rope connecting the sleds is 120.0 N.
c) To determine how far the sleds will travel after the student lets them go, we need to find the distance covered during the 3.00 s while the student was pulling the sleds. We can use the equation:
Distance = (1/2) * acceleration * time^2
Plugging in the values:
Distance = (1/2) * 1.02 m/s^2 * (3.00 s)^2 = 4.59 m
Therefore, the sleds will travel 4.59 m after the student lets them go.
To solve this problem, we will use Newton's second law of motion, which states that the net force on an object is equal to the product of its mass and acceleration (F = ma). We will also use the equation for frictional force (f = μN), where μ is the coefficient of friction and N is the normal force.
a) Frictional force on the back sled:
The mass of the back sled is 55.0 kg, and the sled accelerates with a magnitude of 1.02 m/s^2. Using Newton's second law, we can find the net force acting on the back sled:
Net force = mass x acceleration = 55.0 kg x 1.02 m/s^2 = 56.1 N
Since the only force acting on the back sled is friction, we can conclude that the frictional force on the back sled is 56.1 N.
b) Tension in the rope connecting the sleds:
Since the front sled experiences a frictional force of magnitude 58.8 N, the net force acting on the front sled must be 58.8 N. The student exerts a horizontal force of magnitude 230 N on the sleds. The tension in the rope is equal to this force minus the frictional force on the front sled:
Tension = Force applied - Frictional force
Tension = 230 N - 58.8 N = 171.2 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 the sleds in the horizontal direction. Therefore, the only force that causes the sleds to move is the tension in the rope. Using Newton's second law, we can find the acceleration of the sleds:
Net force = Tension = mass x acceleration
171.2 N = (55.0 kg + 60.0 kg) x acceleration
171.2 N = 115.0 kg x acceleration
acceleration = 1.49 m/s^2
Since the sleds start from rest, we can use the equation of motion s = ut + 0.5at^2 to find the distance traveled by the sleds.
Initial velocity (u) = 0 m/s
Time (t) = 3.00 s
Acceleration (a) = 1.49 m/s^2
Using this equation, we can find the distance (s):
s = 0 + 0.5 x 1.49 m/s^2 x (3.00 s)^2
s = 3.36 m
Therefore, the sleds will travel a distance of 3.36 meters after the student lets them go.
To solve this problem, we need to use Newton's second law of motion and apply it to both sleds individually. Let's go through each step to find the answers to the given questions.
a) To find the frictional force on the back sled, we need to use the equation of motion:
F_net = m*a
where F_net is the net force acting on the sled, m is the mass of the sled, and a is its acceleration.
Since the sled is accelerating, we know that there must be a net force acting on it. In this case, the only force acting on the back sled is the frictional force. Therefore:
F_net = fr = m*a
where fr is the frictional force on the back sled, m is the mass of the sled, and a is its acceleration.
Given that the mass of the back sled is 55.0 kg and the acceleration is 1.02 m/s^2, we can substitute these values into the equation:
fr = 55.0 kg * (1.02 m/s^2)
fr = 56.1 N
So the frictional force on the back sled is 56.1 N.
b) To find the tension in the rope connecting the sleds, we need to consider the net force acting on the front sled. The tension in the rope is responsible for providing the force to move the sled forward.
F_net = T - fr
where F_net is the net force acting on the front sled, T is the tension in the rope, and fr is the frictional force on the front sled.
Given that the frictional force on the front sled is 58.8 N, we can substitute this value into the equation:
230 N = T - 58.8 N
T = 288.8 N
So the tension in the rope connecting the sleds is 288.8 N.
c) To find how far the sleds will travel after they are let go, we need to use the equation of motion:
d = v_0*t + (1/2)*a*t^2
where d is the distance traveled, v_0 is the initial velocity (which is 0 in this case since the sleds start from rest), a is the acceleration, and t is the time.
Given that the sleds are let go after 3.00 s and the acceleration is 1.02 m/s^2, we can substitute these values into the equation:
d = 0 + (1/2)*(1.02 m/s^2)*(3.00 s)^2
d = 0 + 4.59 m
d = 4.59 m
So the sleds will travel a distance of 4.59 meters after they are let go.
I hope this explanation helps! Let me know if you have any further questions.