Suppose a child is riding on a merry-go-round at a distance about halfway between its center and edge. She has a lunch box resting on wax paper, so that there is very little friction between it and the merry-go-round. Which path shown in Figure will the lunch box take when she lets go? The lunch box leaves a trail in the dust on the merry-go-round. Is that trail straight, curved to the left, or curved to the right? Explain your answer.

A force is an influence that can alter an object's motion. An object with mass can change its velocity, or accelerate, as a result of a force. Intuitively, force can be described as a push or a pull. A force is a vector quantity since it has both magnitude and direction.

According to Newton’s second Law of motion the net force is$\mathrm{F}=\mathrm{mXa}$

Here m is the mass and a is acceleration.

The force which is required to keep the object in the circular path is called as centripetal force.The directions of the centripetal force is directed towards the centre of the circle.

The Centripetal force is,$\mathrm{F}=\frac{{\mathrm{mv}}^2}{\mathrm{r}}$

Here$\mathrm{m}$is the mass,$\mathrm{r}$is the distance, and$\mathrm{v}$is the velocity.

In this situation, there is no force. There is very little friction between lunch box and merry–go–round. Hence the only centripetal force will act on the lunch box.

Centripetal force is towards the centre . Therefore, the lunch box will tend to continue straightforward.

Therefore, the lunch will follow path B.

The Lunch will continue straightforwardly.

After the lunch box leaves a trail in the dust in the merry-go-round, there will be no forces acting on the lunch box.

According to Newton’s Second Law of motion, the net force is $\mathrm{F}=\mathrm{mXa}$,

substitute $0$for $\mathrm{F},0=\mathrm{mXa}$,$\mathrm{a}=0$

Therefore, there will be no acceleration on the box.

Hence, the trial will be straightforward.