Acceleration relative to ice

[foreva_lynn]

On a cold winter morning, a child sits on a sled resting on smooth ice. When the 9.50kg sled is pulled with a horizontal force of 31.0 N, it begins to move with an acceleration of 2.50 m/s^2.
The 24.0kg 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. what is the acceleration of the child relative to the ice?

i dnt even know where to begin let alone attempt the problem. please help me out. and tell me the answer so that wen i calculate i will know what my answer will turn out like. thanks very much.

 

[Doc Al]

What must be the force between sled and child? Hint: Apply Newton's 2nd law to the sled.

 

[tomcue]

Hello,

I am happy to provide you with an explanation of the situation and help you solve the problem.

First, let's define what acceleration means. Acceleration is the rate of change of an object's velocity over time. In simpler terms, it is how fast an object is speeding up or slowing down.

In this scenario, the child and the sled are both accelerating, but with different rates. The sled has an acceleration of 2.50 m/s^2, which means that every second, its velocity increases by 2.50 m/s. On the other hand, the child's acceleration is not given, but we know that it is smaller than the sled's acceleration.

Now, let's consider the child's motion relative to the ice. Relative motion means the motion of one object with respect to another. In this case, the child's motion is relative to the ice because the child is sitting on the sled, which is on the ice.

Since we know the sled's acceleration and the child's acceleration relative to the sled, we can use the concept of relative motion to find the child's acceleration relative to the ice.

We can use the following formula to calculate relative acceleration: Relative acceleration = Acceleration of object 1 - Acceleration of object 2

In this case, object 1 is the sled, and object 2 is the child. So, the relative acceleration of the child with respect to the ice would be:

Relative acceleration = 2.50 m/s^2 - Acceleration of the child

We know that the sled's acceleration is 2.50 m/s^2, and the mass of the sled is 9.50 kg. We can use Newton's second law of motion (F=ma) to find the force acting on the sled:

31.0 N = (9.50 kg) x (2.50 m/s^2)

This gives us the force acting on the sled, which is 31.0 N. Since the sled and the child are connected, they experience the same force. Therefore, the force acting on the child is also 31.0 N.

Now, we can use Newton's second law again to find the child's acceleration:

31.0 N = (24.0 kg) x (Acceleration of the child)

Solving for the acceleration of the child, we get:

Acceleration of the child = 31.0 N / 24.0