Two fi gure skaters are moving east together during a
performance. Skater 1 (78 kg) is behind skater 2 (56 kg)
when skater 2 pushes on skater 1 with a force of
64 N [W]. Assume that no friction acts on either
skater. T / I
(a) Determine the acceleration of each skater.
(b) What will happen to the motion of each skater?
Explain your reasoning
Fnet = Ft
Skater 1: a = Ft/m
a = 64N/78kg
a = 0.82m/s2[E] kg
Skater 2: a = Ft/m
a = 64N/56kg
a = 1.1 m/s2 [W]
The skaters will accelerate in the opposite direction due to how the action-reaction law states that for every action force there is a simultaneous reaction force equal in magnitude but opposite in direction.
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(a) Well, let's tackle the first part of the question: the acceleration of each skater. We can use Newton's second law, which states that force is equal to mass times acceleration (F = ma).
Skater 1 has a mass of 78 kg and experiences a force of 64 N to the west. So we can use F = ma to find the acceleration of Skater 1: 64 N = 78 kg * a1. Solving for a1, we find that a1 = 64 N / 78 kg.
For Skater 2, the force being applied is the same but in the opposite direction (east). Using F = ma again, we have 64 N = 56 kg * a2. Solving for a2, we find that a2 = 64 N / 56 kg.
(b) Now, let's think about what will happen to the motion of each skater. Since Skater 1 experiences a force to the west and has a greater mass, their acceleration (a1) will be less than the acceleration (a2) of Skater 2, who experiences the same force in the opposite direction.
As a result, Skater 2 will have a greater magnitude of acceleration than Skater 1, meaning Skater 2 will move faster than Skater 1. Skater 2 will pull away from Skater 1, increasing the distance between them.
So in summary, Skater 1 will have a smaller acceleration and Skater 2 will have a larger acceleration. Skater 2 will move ahead of Skater 1, creating some distance between them. Looks like Skater 2 is taking the lead in this performance!
To determine the acceleration of each skater, we can 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.
(a) Acceleration of Skater 1 (m1 = 78 kg):
The net force acting on Skater 1 can be calculated by subtracting the force exerted by Skater 2 from the force exerted by Skater 1.
Net force on Skater 1 = Force exerted by Skater 1 - Force exerted by Skater 2
Given:
Force exerted by Skater 1 = 64 N [W]
Force exerted by Skater 2 = 64 N [W] (equal in magnitude but opposite in direction)
Net force on Skater 1 = 64 N [W] - 64 N [W] = 0 N
Since there is no net force acting on Skater 1, their acceleration will be zero (0 m/s²).
Acceleration of Skater 1 (a1) = 0 m/s²
Acceleration of Skater 2 (m2 = 56 kg):
The net force acting on Skater 2 is equal to the force exerted by Skater 2 (as there are no other forces acting on Skater 2).
Net force on Skater 2 = Force exerted by Skater 2 = 64 N [W]
Using Newton's second law, we can calculate the acceleration of Skater 2:
Net force = m * a
64 N [W] = 56 kg * a2
Dividing both sides of the equation by the mass of Skater 2:
a2 = 64 N [W] / 56 kg ≈ 1.143 m/s²
Acceleration of Skater 2 (a2) ≈ 1.143 m/s²
(b) Based on the calculated accelerations:
Since Skater 1 has an acceleration of 0 m/s², it means that there is no change in their velocity. They will continue to move at a constant velocity without any change in motion.
On the other hand, Skater 2 has a non-zero acceleration of approximately 1.143 m/s². This means that there is a net force acting on Skater 2, causing them to accelerate in the westward direction. Skater 2 will experience a change in motion and their velocity will gradually increase towards the west.
(a) To determine the acceleration of each skater, we need to apply Newton's second law of motion, which states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. Mathematically, it can be expressed as:
Acceleration = Net force / Mass
For skater 1:
Mass of skater 1 (m1) = 78 kg
Net force on skater 1 (F1) = 64 N [W]
Acceleration of skater 1 (a1) = F1 / m1
= 64 N / 78 kg
For skater 2:
Mass of skater 2 (m2) = 56 kg
Net force on skater 2 (F2) = -64 N [E] (opposite direction to skater 1's force, due to Newton's third law)
Acceleration of skater 2 (a2) = F2 / m2
= -64 N / 56 kg
So, the acceleration of skater 1 is approximately 0.821 m/s^2 [W] and the acceleration of skater 2 is approximately -1.143 m/s^2 [E].
(b) The negative acceleration of skater 2 indicates that they will experience a deceleration in the opposite direction of the applied force. This means that skater 2 will slow down.
As for skater 1, they will experience an acceleration in the direction of the applied force, which means they will speed up.
However, since skater 2 pushes on skater 1, it will cause skater 1 to move forward slightly while skater 2 moves backward. The overall effect will be that both skaters will slow down but with different rates.
In summary, skater 2 will slow down more quickly due to the greater force and smaller mass, while skater 1 will slow down at a slower rate.