High school Physics Acceleration

Question

A 50kg bobsled slides down an ice track starting (at zero initial speed) from the top of a 148m high hill. The acceleration of gravity is 9.8 m/s2 . Neglect friction and air resistance and determine the bobsled’s speed at the bottom of the hill.

Answer 1

I'm going to assume that you are going through a section on the conservation of energy.

The total energy (E) of a system like this which contains gravitational potential energy (E_P) and kinetic energy (E_K) is the sum of those. E_TOTAL = E_P + E_K

Remember that gravitational potential energy (E_P) is defined as E_P = mgh and kinetic energy (E_K) is defined as E_K = 1/2mv².

At time "1", the bobsled is at the top of the hill and has zero speed. So the total energy is defined as

E_TOTAL-1 = E_P1 + E_K1 = mgh + 1/2mv² = (70)(9.8)(148) + 1/2(70)(0)² = 101528 + 0 = 101528 joules

At time "2", the bobsled is at the bottom of the hill and has an unknown speed (v). So the total energy is defined as

E_TOTAL-2 = E_P2 + E_K2 = mgh + 1/2mv² = (70)(9.8)(0) + 1/2(70)(v)² = 0 + 35v² = 35v²

Since energy is conserved, then E_TOTAL-1 = E_TOTAL-2 therefore:

101528 = 35v²

2900.8 = v²

v = √2900.8

v = 53.859

or v = 54 m/s

Answer 2

By Law of Conservation of Energy: mgh = mv²/2; v = √2gh = √2·9.8m/s²·148m = 53.86 m/s