Question

A bobsled zips down an ice track, starting from rest at the top of a hill with a vertical height of 150m. Disregarding friction, what is the velocity of the bobsled at the bottom of the hill? (g=9.81 m/s^2)

Answer 1

The velocity would be 54.2 m/s We would use the equation 1/2mv^2top+mghtop = 1/2mv^2bottom+mghbottom where m is the mass of the bobsled(which can be ignored), vtop/bottom is the velocity of the bobsled at the top or bottom, g is gravity, and htop/bottom is the height of the bobsled at the top or bottom of the hill. Since the velocity of the bobsled at the top of the hill and height at the bottom of the hill are zero, 1/2mv^2top and mghbottom will equal zero. The equation will be mghtop=1/2mv^2bottom. Thus we would solve for v.

Answer 2

Answer:

54.2 m/s

Explanation:

We can solve the problem by using the law of conservation of energy. The total mechanical energy at the top of the hill is just gravitational potential energy:

$E_i = mgh$

where m is the mass of the bobsled, g is the gravitational acceleration and h=150 m. The total mechanical energy at the bottom of the hill is just kinetic energy:

$E_f = \frac{1}{2}mv^2$

where v is the velocity of the bobsled at the bottom of the hill. Since the total energy must be conserved, we can write:

$mgh=\frac{1}{2}mv^2$

and we can solve for v

$v=\sqrt{2gh}=\sqrt{2(9.81 m/s^2)(150 m)}=54.2 m/s$