A student has tied together two sleds with a rope. Two of his friends, with masses of 55.0 kg and 60.0

Question

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/s2
. 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 after he lets them go?

Answer 1

a) To find the frictional force on the back sled, we can use Newton's second law:
F = ma
Where F is the net force on the object, m is the mass of the object, and a is the acceleration of the object.

Since the only force acting on the back sled is the tension in the rope, and we know the magnitude of the tension (230 N), we can use this force in our equation:

F = ma
230 N - f = (55.0 kg)(1.02 m/s^2)
230 N - f = 56.1 N
f = 230 N - 56.1 N
f = 173.9 N

Therefore, the frictional force on the back sled is 173.9 N.

b) To find the tension in the rope connecting the sleds, we can use the same equation as before:
F = ma

The net force on the front sled is the tension in the rope minus the frictional force:
230 N - 58.8 N = (60.0 kg)(1.02 m/s^2)
171.2 N = 61.2 N

Therefore, the tension in the rope connecting the sleds is 171.2 N.

c) After the student lets the sleds go, they will continue to move at a constant velocity due to inertia. The distance they travel can be calculated using the equation:
d = v * t

Where d is the distance, v is the velocity, and t is the time.

Since the sleds are moving with a constant velocity, the distance they travel after 3.00 s is:
d = (1.02 m/s)(3.00 s)
d = 3.06 m

Therefore, the sleds will travel a distance of 3.06 meters after the student lets them go.