an electric motor is used to pull a 125 kg box across a floor using a long cable. the tension in the cable is 350 N amd the box accelerates at 1.2 m/s^2 [forward] for 5 seconds. the cable breaks and the box slows down and stops. calculate the coefficient of kinetic friction? How far does the box travel up to the moment the cable breaks? how far does the box travel from the moment the cable breaks until it stops?
See your previous post for solution.
To calculate the coefficient of kinetic friction, we need to first determine the net force acting on the box while it is being pulled forward by the electric motor.
The net force can be calculated using Newton's second law of motion:
Net Force = Mass x Acceleration
Given:
- Mass of box (m) = 125 kg
- Acceleration (a) = 1.2 m/s^2
Substituting the values, the net force is:
Net Force = 125 kg x 1.2 m/s^2 = 150 N
Since the tension in the cable (T) is 350 N and the cable breaks, the tension is no longer present. Hence, the force acting opposite to the direction of motion is the force of kinetic friction (Fk).
We can now calculate the coefficient of kinetic friction (μk) using the equation:
Fk = μk x Normal Force
The normal force (N) is equal to the weight of the box, which is the mass (m) multiplied by the acceleration due to gravity (g). Assuming g to be 9.8 m/s^2:
Normal Force = m x g
Normal Force = 125 kg x 9.8 m/s^2
Normal Force = 1225 N
Since the box is moving horizontally, the normal force is equal to the force of gravity acting vertically downward.
Substituting the values into the equation for kinetic friction:
150 N = μk x 1225 N
Now, solve for the coefficient of kinetic friction (μk):
μk = 150 N / 1225 N ≈ 0.122
Therefore, the coefficient of kinetic friction is approximately 0.122.
To calculate the distance traveled by the box up to the moment the cable breaks:
The distance traveled is given by the equation:
Distance = Initial velocity x Time + 0.5 x Acceleration x Time^2
Given:
- Initial velocity (v0) = 0 (as the box starts from rest)
- Acceleration (a) = 1.2 m/s^2
- Time (t) = 5 seconds
Substituting the values:
Distance = 0 x 5 + 0.5 x 1.2 m/s^2 x (5 seconds)^2
Distance = 0 + 0.5 x 1.2 m/s^2 x 25 seconds^2
Distance = 0 + 0.5 x 1.2 m/s^2 x 625 seconds^2
Distance ≈ 375 meters
Therefore, the box travels approximately 375 meters up to the moment the cable breaks.
To calculate the distance traveled by the box from the moment the cable breaks until it stops:
Since the cable breaks, the net force acting on the box is the force of kinetic friction (Fk).
Using the equation:
Force = Mass x Acceleration
Given:
- Force (F) = Fk (force of kinetic friction) = 150 N
- Mass (m) = 125 kg
Substituting the values into the equation:
Fk = 125 kg x Acceleration
Rearranging the equation to solve for acceleration:
Acceleration = Fk / m
Acceleration = 150 N / 125 kg
Acceleration = 1.2 m/s^2
Now, we can use the equation of motion to calculate the distance:
Distance = (Final velocity)^2 - (Initial velocity)^2 / 2 x Acceleration
Assuming the final velocity (vf) is 0 (because the box stops), and the initial velocity (vi) is unknown, we can rewrite the equation as:
Distance = - vi^2 / (2 x Acceleration)
Solving for the distance:
Distance = - vi^2 / (2 x 1.2 m/s^2)
0 = - vi^2 / 2.4 m/s^2
Since the initial velocity square cannot be negative, it means the box comes to a stop but does not travel any additional distance after the cable breaks.
Therefore, the distance traveled by the box from the moment the cable breaks until it stops is 0 meters.