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/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?
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.
To solve the problem, we can apply Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.
a) Frictional force on the back sled:
The net force acting on the back sled is the tension in the rope pulling it forward minus the frictional force:
Net force = Tension - Friction
Tension = Net force + Friction
Now, we can calculate the tension using the given values:
Net force = mass × acceleration
Net force = 55.0 kg × 1.02 m/s^2
Net force = 56.1 N
Tension = 56.1 N + 58.8 N
Tension = 114.9 N
b) Tension in the rope connecting the sleds:
The tension in the rope connecting the sleds is equal to the tension being applied by the student:
Tension = 230 N
c) Distance traveled by the sleds after the student lets them go:
To find the distance traveled, we need to calculate the final velocity of the sleds when the student lets them go. We can use the equation of motion:
v^2 = u^2 + 2as
Where:
v = final velocity (unknown)
u = initial velocity (0, as the student starts from rest)
a = acceleration (1.02 m/s^2)
s = distance (unknown)
Rearranging the equation:
v^2 = 0^2 + 2(1.02 m/s^2)s
v^2 = 2.04s
Substituting the given values:
(1.02 m/s^2)s = v^2
Now, we will use the third equation of motion:
v = u + at
Where:
v = final velocity (unknown)
u = initial velocity (0, as the student starts from rest)
a = acceleration (1.02 m/s^2)
t = time (3.00 s)
Substituting the values:
v = 0 + (1.02 m/s^2)(3.00 s)
v = 3.06 m/s
Now, substitute this value of v into the first equation:
(1.02 m/s^2)s = (3.06 m/s)^2
s = (3.06 m/s)^2 / (1.02 m/s^2)
s = 9.3636 m
Therefore, the sleds will travel approximately 9.36 meters 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 acting on an object is equal to the product of its mass and acceleration (F = ma). We'll also consider the concepts of friction and tension.
a) To find the frictional force on the back sled, we need to apply Newton's second law to the back sled. The net force on the back sled is the tension in the rope connecting the sleds minus the frictional force:
Net force = Tension - Frictional force
We know the tension in the rope connecting the sleds is equal to the force applied by the student, which is 230 N. The net force on the back sled is given by the mass of the back friend multiplied by the acceleration:
Net force = Mass of back friend * Acceleration
Setting both expressions for the net force equal, we have:
Tension - Frictional force = Mass of back friend * Acceleration
Rearranging the equation, we can solve for the frictional force:
Frictional force = Tension - Mass of back friend * Acceleration
Substituting the given values:
Frictional force = 230 N - (55.0 kg * 1.02 m/s^2)
Calculating the frictional force gives:
Frictional force = 230 N - 56.1 N = 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 already know that it is equal to 230 N, as given in the problem.
Therefore, the tension in the rope connecting the sleds is 230 N.
c) To determine how far the sleds will travel after the student lets them go, we need to calculate the distance traveled in the 3-second interval when the student is pulling the sleds. To do this, we can use the equations of motion.
The equation for distance traveled in terms of initial velocity, acceleration, and time is:
Distance = (Initial velocity * Time) + (1/2 * Acceleration * Time^2)
In this case, the initial velocity is 0 m/s since the student starts from rest, and the acceleration is 1.02 m/s^2. Plugging in these values and the time of 3.00 s gives us:
Distance = (0 m/s * 3.00 s) + (1/2 * 1.02 m/s^2 * (3.00 s)^2)
Simplifying the equation, we find:
Distance = 0 m + (1/2 * 1.02 m/s^2 * 9.00 s^2) = 4.59 m
Therefore, the sleds will travel a distance of 4.59 meters after the student lets them go.