# A block of mass m starts from rest at the top of a friction-less hill of height H. Its slides...

## Question:

A block of mass m starts from rest at the top of a friction-less hill of height H. Its slides down the hill and then over another hill of height h. The top of the second smaller hill can be modeled as part of a circle with radius h.

If H = 5h/4, what is the normal force that the track exerts on the block when it is at the top of the second hill? Express your answer in terms of m and g. If any part of your solution involves a free-body diagram, it must be clearly shown.

## Energy conservation

We know that from the law of energy conservation that energy can neither be created nor be destroyed it can only change its forms from one to another. Therefore, the total energy of the system would remain constant at every instant.

The initial energy of the mass m is given by

{eq}\displaystyle E_{i} = mgH {/eq}

When the mass m has reached to the top of the other hill

{eq}\displaystyle E_{f} = mgh + 0.5mv^{2} {/eq}

Now applying the energy conservation

{eq}\displaystyle E_{i} = E_{f} \\ mgH = mgh + 0.5mv^{2} \\ mg\left(\frac{5h}{4}\right) = mgh + 0.5mv^{2} \\ v^{2} = \frac{gh}{2} -----------(1) {/eq}

Now applying Newton's second law at the top of the smaller hill

{eq}\displaystyle mg - N = \frac{mv^{2}}{h} \\ mg - N = \frac{m*gh}{2h} \\ N = mg - \frac{mg}{2} \\ N = \frac{mg}{2} {/eq}

Where

• N is normal reaction
• h is also the radius of the circular loop