Unit 1 Question 3

Question

A sled and rider with a combined weight of 60 kg are at rest on the top of the hill of 12 m high. Assuming there is no friction along the slope, calculate the speed at which the sled will be moving when it will reach the bottom of the hill.

Answer 1

To calculate the speed at which the sled and rider will be moving when they reach the bottom of the hill, you can use the principles of conservation of energy. At the top of the hill, the sled and rider have potential energy, and as they move downhill, this potential energy is converted into kinetic energy. We can equate the two forms of energy as follows:

Potential Energy (at the top) = Kinetic Energy (at the bottom)

The potential energy at the top of the hill is given by:

PE = m * g * h

Where:

PE is the potential energy.

m is the mass (60 kg).

g is the acceleration due to gravity (approximately 9.81 m/s²).

h is the height of the hill (12 m).

So,

PE = 60 kg * 9.81 m/s² * 12 m

PE = 7069.2 J (joules)

Now, at the bottom of the hill, this potential energy is converted into kinetic energy:

KE = (1/2) * m * v^2

Where:

KE is the kinetic energy.

m is the mass (60 kg).

v is the velocity (speed) at the bottom.

We can set the potential energy at the top equal to the kinetic energy at the bottom:

PE = KE

So,

7069.2 J = (1/2) * 60 kg * v^2

Now, solve for v:

v^2 = (2 * 7069.2 J) / (60 kg)

v^2 = (2 * 7069.2) / 60

v^2 = 235.64

v ≈ √235.64

v ≈ 15.35 m/s

So, the speed at which the sled and rider will be moving when they reach the bottom of the hill is approximately 15.35 meters per second.

Answer 2

It's a conservation of energy problem.

Final kinetic energy (KE = ½mv²) at the bottom is equal to the initial potential energy at the top (PE = mgh).

So ½mv² = mgh. Divide both sides by m to get

½v² = gh, which means that v² = 2gh and v =√(2gh)

Now plug in h = 12 m and g = 9.8 m/s² to get

v =√[2(9.8)(12)] ≈ 15.3 m/s ≈ 15 m/s (How you round off might depend on getting significant figures correct.)