Question

Shane goes roller skating. Shane has a mass of 65kg. He rolls down a hill that is 10m high. He then rolls 13m up the following hill before coming to a rest. How fast was he rolling at the top of the first hill?

Answer 1

To determine the speed at the top of the first hill, we can use the principle of conservation of energy. At the top of the hill, Shane's gravitational potential energy is converted into kinetic energy.

Step-by-step explanation:

To determine the speed at the top of the first hill, we can use the principle of conservation of energy. At the top of the hill, Shane's gravitational potential energy is converted into kinetic energy. Therefore, we can equate the gravitational potential energy at the top of the first hill with the kinetic energy at that point.

Gravitational potential energy = mgh

Kinetic energy = (1/2)mv^2

Setting them equal to each other, we get:

mgh = (1/2)mv^2

Where:

• m = mass of Shane (65 kg)
• g = acceleration due to gravity (9.8 m/s^2)
• h = height of the first hill (10 m)
• v = velocity at the top of the first hill (to be determined)

Plugging in the values, we get:

(65 kg)(9.8 m/s^2)(10 m) = (1/2)(65 kg)v^2

Simplifying the equation gives:

637 m = (1/2)v^2

Dividing both sides by (1/2) gives:

v^2 = 1274

Taking the square root of both sides gives:

v ≈ 35.7 m/s

Therefore, Shane was rolling at approximately 35.7 m/s at the top of the first hill.