On a cold winter morning, a child sits on a sled resting on smooth ice. When the 9.20 kg sled is pulled with a horizontal force of 40.0 N, it begins to move with an acceleration of 2.50 m/s^2.
The 23.0 kg child accelerates too, but with a smaller acceleration than that of the sled. Thus, the child moves forward relative to the ice, but slides backward relative to the sled.Find the acceleration of the child relative to the ice.
To find the acceleration of the child relative to the ice, we need to find the net force acting on the child and divide it by the child's mass.
Let's break down the forces acting on the child:
1. Force of friction between the child and the sled: This force opposes the motion of the child and acts in the direction opposite to the force applied on the sled.
2. Force of friction between the sled and the ice: This force opposes the motion of the sled and acts in the direction opposite to the force applied on the sled.
Since the child is moving backward relative to the sled, we can assume that the force of friction between the child and the sled is greater than the applied force.
Now, let's use Newton's second law of motion (F = ma) to find the net force acting on the child:
Net force = Force applied on the sled - Force of friction between the child and the sled
Given:
Mass of the sled (m_s) = 9.20 kg
Force applied on the sled (F_applied) = 40.0 N
Acceleration of the sled (a_sled) = 2.50 m/s^2
Mass of the child (m_c) = 23.0 kg
Net force = F_applied - Force of friction between the child and the sled
To find the force of friction between the child and the sled, we can use the equation:
Force of friction = mass of the child x acceleration of the sled
Force of friction (F_friction) = m_c x a_sled
F_friction = 23.0 kg x 2.50 m/s^2
F_friction = 57.5 N
Now, we can find the net force acting on the child:
Net force = F_applied - F_friction
Net force = 40.0 N - 57.5 N
Net force = -17.5 N (since it acts in the opposite direction to the applied force)
Finally, we can find the acceleration of the child relative to the ice using Newton's second law:
Net force = mass of the child x acceleration of the child
-17.5 N = 23.0 kg x acceleration of the child
Acceleration of the child (a_child) = -17.5 N / 23.0 kg
a_child = -0.76 m/s²
So, the acceleration of the child relative to the ice is -0.76 m/s² (backward relative to the sled).
To find the acceleration of the child relative to the ice, 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.
Let's designate the acceleration of the child relative to the ice as "a_child."
The total mass of the system (child + sled) is the sum of the masses of the child and the sled: m_total = 9.20 kg + 23.0 kg = 32.2 kg.
Since the sled and child are connected, the force applied to the sled is equal to the force applied to the child. Therefore, the force acting on the child is 40.0 N.
The total force acting on the system is the sum of the forces acting on the sled and the child. The force acting on the sled causes its acceleration, so we can use the sled's mass and acceleration to calculate this force:
F_sled = m_sled x a_sled = 9.20 kg x 2.50 m/s^2 = 23.0 N.
Since the forces are equal, the force acting on the child is also 23.0 N.
Now we can use Newton's second law to calculate the acceleration of the child:
F_child = m_child x a_child
23.0 N = 23.0 kg x a_child
Dividing both sides by 23.0 kg, we find:
a_child = 23.0 N / 23.0 kg = 1.00 m/s^2.
Therefore, the acceleration of the child relative to the ice is 1.00 m/s^2.