2 people, each with a mass of 70 kg, are wearing inline skates and are holding opposite ends of a 15 m rope. One person pulls forward on the rope by moving hand over hand and gradually reeling in more of the rope. In doing so, he exerts a force of 35 N [backwards] on the rope. This causes him to accelerate toward the other person. Assuming that the friction acting on the skaters is negligible, how long will it take for them to meet?
To find the time it takes for the two people to meet, we can use Newton's second law of motion.
1. Determine the net force acting on the system:
- The only force acting on the system is the force exerted by the person pulling the rope, which is 35 N in the backward direction.
- Since the two people are in opposite directions, the net force acting on the system is the difference between the force exerted by the person pulling the rope and the force exerted by the person standing still, which is equal in magnitude but opposite in direction. Therefore, the net force is 35 N - (-35 N) = 70 N in the forward direction.
2. Use Newton's second law to find acceleration:
- Newton's second law states that the acceleration of an object is equal to the net force acting on it divided by its mass. In this case, the mass of each person is 70 kg.
- Therefore, the acceleration of the system is 70 N / (70 kg + 70 kg) = 0.5 m/s².
3. Find the distance traveled by the person pulling the rope:
- The distance traveled by the person pulling the rope can be calculated using the equation of motion: s = ut + (1/2)at², where s is the distance, u is the initial velocity (which is 0 in this case), t is the time, and a is the acceleration.
- The distance traveled is equal to the length of the rope, which is 15 m.
- Since the initial velocity is 0, the equation simplifies to s = (1/2)at².
- Plugging in the values, we get 15 m = (1/2)(0.5 m/s²)(t²).
- Rearranging the equation, we have t² = (15 m) / (0.5 m/s²) = 30 s².
4. Solve for time:
- Taking the square root of both sides of the equation, we find t ≈ √30 ≈ 5.5 s.
Therefore, it will take approximately 5.5 seconds for the two people to meet.
To find the time it takes for the two people to meet, we can use 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.
In this scenario, the only force acting on the system is the force exerted by the person pulling backward on the rope. This force causes both individuals to accelerate toward each other.
First, let's calculate the acceleration of the system using Newton's second law:
F = m * a
Where F is the net force, m is the total mass of the two individuals, and a is the acceleration of the system.
In this problem, the force (F) is given as 35 N. The total mass (m) is the sum of the masses of the two individuals, which is 70 kg + 70 kg = 140 kg.
Therefore:
35 N = 140 kg * a
Simplifying the equation:
a = 35 N / 140 kg
a = 0.25 m/s²
Now that we have the acceleration, we can find the time it takes for the two individuals to meet using the kinematic equation:
Δx = v₀t + 0.5at²
Where Δx is the distance traveled, v₀ is the initial velocity, a is the acceleration, t is the time, and t² represents t squared.
In this scenario, the initial velocity is 0 because both individuals start from rest. The distance traveled (Δx) is given as 15 m, and the acceleration (a) is 0.25 m/s².
Plugging in the values:
15 m = 0.5 * (0.25 m/s²) * t²
Simplifying further:
15 m = 0.125 m/s² * t²
Rearranging the equation to solve for time squared:
t² = (15 m) / (0.125 m/s²)
t² = 120 s²
Finally, taking the square root of both sides to solve for time:
t = √(120 s²)
t ≈ 10.95 s
Therefore, it will take approximately 10.95 seconds for the two individuals to meet.