Question

A Sunset student, whose mass is 60 kg, is standing at the top of a 50-meter hill on a snowy day. The student slips on some ice and slides to the bus stop, which is a vertical drop of 8 meters. What is the student's speed when they reach the bus stop at the bottom of the hill?

A) 8 m/s
B) 10 m/s
C) 12 m/s
D) 15 m/s

Answer 1

The conservation of mechanical energy is used to calculate the speed of a Sunset student who slides down a hill with a vertical drop of 8 meters. By solving the energy conservation equation, the student's speed is found to be approximately 12 m/s when they reach the bus stop at the bottom of the hill.

Step-by-step explanation:

The question is asking to calculate the student's speed when they reach the bus stop at the bottom of a hill with a vertical drop of 8 meters. Since the question provides the mass of the student and the vertical drop, and advises to ignore any friction or air resistance, this is a classic example of the conservation of mechanical energy. The mechanical energy at the start (potential energy at the top of the hill) is equal to the mechanical energy at the end (kinetic energy at the bottom).

The potential energy (PE) at the top of the hill is given by PE = mgh, where m is mass, g is the acceleration due to gravity (9.8 m/s2), and h is height. The kinetic energy (KE) at the bottom is given by KE = 0.5 * m * v2, where v is velocity.

Setting PE equal to KE, because of conservation of energy, and solving for v, we get:

1. PE = mgh = 60 kg * 9.8 m/s2 * 8 m
2. KE at the bottom = PE at the top
3. 0.5 * m * v2 = 60 kg * 9.8 m/s2 * 8 m
4. Solving for v gives v = sqrt((2 * m * g * h) / m)
5. v = sqrt(2 * 9.8 m/s2 * 8 m)
6. v ≈ 12.6 m/s

However, since this answer is not one of the provided options, we might have to consider that the possible answers are approximations and choose the closest one, which is 12 m/s, option C.