4. An ice skater pulls three small children, one behind the other, with masses 25kg, 31 kg and 35kg. Assume that the ice is smooth enough to be considered frictionless, determine:
a) the total force applied to the train of children if they reach a speed of 3.5m/s from rest in 15 s;
b) if the skater is holding onto the 25kg child, find the tension in the arms of the next child (31 kg) in line.
To determine the total force applied to the train of children, 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.
a) To find the total force applied to the train of children, we first need to find the total mass of the children. We can do this by adding up their individual masses:
Total mass (m_total) = mass_child1 + mass_child2 + mass_child3
= 25 kg + 31 kg + 35 kg
= 91 kg
Next, we need to find the acceleration of the train of children. We can use the equation of motion: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
Given:
Initial velocity (u) = 0 m/s
Final velocity (v) = 3.5 m/s
Time (t) = 15 s
Using the equation, we can solve for acceleration (a):
v = u + at
3.5 m/s = 0 m/s + a * 15 s
3.5 m/s = 15 a
a = 3.5 m/s / 15 s
a ≈ 0.2333 m/s²
Now, we can find the total force by multiplying the total mass by the acceleration:
Total force (F_total) = m_total * a
= 91 kg * 0.2333 m/s²
≈ 21.25 N
Therefore, the total force applied to the train of children is approximately 21.25 N.
b) To find the tension in the arms of the next child (31 kg) in line, we need to consider the forces acting on this child. The child is being pulled forward by the skater and backward by the third child. We neglect any friction between the children.
The tension in the arms of the next child is equal to the force experienced by the child in the backward direction.
Since the ice is frictionless, the force exerted by the skater on the first child is equal to the force exerted by the third child on the second child. Therefore, the tension in the arms of the next child is equal to the force exerted by the third child on the second child.
Tension in the arms of the next child = Force exerted by the third child on the second child
To find the force exerted by the third child on the second child, we can use Newton's second law again.
Force (F) = mass (m) * acceleration (a)
Given:
Mass of the third child (m_third_child) = 35 kg
Acceleration (a) = 0.2333 m/s² (from the previous calculation)
Force exerted by the third child on the second child = m_third_child * a
= 35 kg * 0.2333 m/s²
≈ 8.166 N
Therefore, the tension in the arms of the next child (31 kg) in line is approximately 8.166 N.
a. a = (3.5 - 0) /(15 - 0)=0.23m/s^2.
F = ma = (25 + 31 + 35) * 0.233,
= 91 * 0.233 = 21.2N.