On a cold winter morning, a child sits on a sled resting on smooth ice. When the 9.10 kg sled is pulled with a horizontal force of 36.0 N, it begins to move with an acceleration of 2.10 m/s^2.

Question

On a cold winter morning, a child sits on a sled resting on smooth ice. When the 9.10 kg sled is pulled with a horizontal force of 36.0 N, it begins to move with an acceleration of 2.10 m/s^2.

The 25.0 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.

Answer 1

break the pulling force in to two components, that which is acting on the sled, and that which is acting on the child.

36=Fs+Fc=9.10*2.10+ 25*a
solve for a.

Answer 2

To find the acceleration of the child relative to the ice, we need to determine the net force acting on the child and then divide it by the child's mass.

First, let's analyze the forces acting on the child. There are two forces at play: the force exerted on the child by the sled, and the force of friction between the child and the ice.

The force exerted on the child by the sled can be calculated using Newton's second law:

Force = mass x acceleration

We know the sled's mass is 9.10 kg and its acceleration is 2.10 m/s^2. By rearranging the equation, we can find the force exerted on the child:

Force = (9.10 kg) x (2.10 m/s^2)
= 19.11 N

The force of friction between the child and the ice opposes the motion of the child. The force of friction can be calculated using the equation:

Frictional force = coefficient of friction x normal force

Since the child is sliding backward relative to the sled, the force of friction acts in the positive direction. Thus, the normal force is equal to the force exerted on the child by the sled, which is 19.11 N.

Now, we need to determine the coefficient of friction. Unfortunately, the problem statement does not provide this value. However, we can proceed with the assumption that the child has a static coefficient of friction. A reasonable estimation for the static coefficient of friction between ice and a person's clothing is around 0.03.

With this information, the force of friction becomes:

Frictional force = (0.03) x (19.11 N)
= 0.57 N

The net force acting on the child is the difference between the force exerted on the child by the sled and the force of friction:

Net force = Force - Frictional force
= 19.11 N - 0.57 N
= 18.54 N

Finally, we divide the net force by the child's mass to find the acceleration of the child relative to the ice:

Acceleration = Net force / mass
= 18.54 N / 25.0 kg
= 0.74 m/s^2

Therefore, the acceleration of the child relative to the ice is 0.74 m/s^2.

On a cold winter morning, a child sits on a sled resting on smooth ice. When the 9.10 kg sled is pulled with a horizontal force of 36.0 N, it begins to move with an acceleration of 2.10 m/s^2.

break the pulling force in to two components, that which is acting on the sled, and that which is acting on the child.

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To find the acceleration of the child relative to the ice, we need to determine the net force acting on the child and then divide it by the child's mass.