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.
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?
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