Chapter 5, Problem 5Q

To determine

To explain:

The decrease in the normal force between the child’s chest and the sled while going over the hill using Newton’s second law.

Answer to Problem 5Q

Solution:

A child on a sled comes flying over the crest of a small hill. The sled does not leave the ground, but the child feels the normal force between the chest and the sled decrease as he goes over the hill because resultant of normal force between sled and weight of child provides centripetal force which is downward.

Explanation of Solution

Formula used.

  $\begin{array}{l}{F}_{net}=\frac{m{v}^2}{r}\\ \text{where }{F}_{net}\text{ net resulting force towards the center of circle}\text{.}\end{array}$

m : mass of body

v : speed of circular motion

r : radius of circular path

Calculation:

  

Free body diagramof child when he is at the top of hill

Let radius $=r$

Centripetal acceleration $=\frac{v^2}{r}$ and it is acting toward the centre of curvature of top of hill.

Apply Newton’s second law of motion.

  $mg-N=m\frac{v^2}{r}$

  $N=m\left(g-\frac{v^2}{r}\right)<mg$

From the above expression, it becomes clear that normal force at the top of hill less than weight of boy. So,the child the normal between his chest and the sled decreases as he goes over the heel.

Conclusion:

A child on a sled comes flying over the crest of a small hill, his sled does not leave the ground, but he feels the normal force between his chest and the sled decrease as he goes over the hill because resultant of normal force between sled and weight of child provide centripetal force which is downward.