An ice skater pulls three small children, one behind the other, with masses 25kg, 31kg, and 35kg. Assume that the ice is smooth enough to be considered frictionless.
If the skater is holding onto the 25kg child, find the tension in the arms of the next child in line.
see the remarks I made on the three blocks.
To find the tension in the arms of the next child in line, we can use Newton's Second Law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration.
In this case, the force pulling the first child forward is the tension in the skater's arms, and the total resistance or force opposing the motion is the sum of the masses of the children behind the first child multiplied by their respective accelerations.
The acceleration of all the children will be the same because they are all connected and moving in a straight line.
Let's say the tension in the skater's arms is T_N (the subscript N corresponds to the next child in line). The total resistance can be calculated as:
Total Resistance = (mass of the second child + mass of the third child) × acceleration
Using Newton's Second Law, we can express this as:
Total Resistance = (31 kg + 35 kg) × acceleration
Now, since the skater is pulling only the 25 kg child, the force that contributes to the total resistance is the tension in the skater's arms minus the force needed to move the 25 kg child. This force can be expressed as:
Force = (mass of the first child) × acceleration
Substituting the given values into the equation, we have:
Force = 25 kg × acceleration
Since the force pulling the first child forward is equal to the total resistance, we can equate the two expressions:
T_N - 25 kg × acceleration = Total Resistance
Now we can substitute the total resistance value:
T_N - 25 kg × acceleration = (31 kg + 35 kg) × acceleration
Simplifying the equation, we get:
T_N = (31 kg + 35 kg) × acceleration + 25 kg × acceleration
Finally, we can calculate the tension:
T_N = 66 kg × acceleration + 25 kg × acceleration
The value of acceleration is not given in the problem, so we need more information to determine the exact tension in the next child's arms.