Question

As shown below, a 100 kg student is compressed 50 cm on a spring with a spring constant of k = 80,000 N/m. He is on top of a 10 m frictionless hill. He then is released from rest. He goes down to the bottom of the hill before sliding up a 30° frictionless hill. a. (8 pts) Find the speed of the student when he reaches the bottom of the hill. b. (9 pts) Find the distance D the student travels up the hill before momentarily stopping. Use whichever method you wish.

Answer 1

Given the mass of the student is m = 100kg, the compression of the spring is x = 50cm = 0.5m, the spring constant is k = 80000N/m, the height of the first hill is h1 = 10m and the slope of the second hill is .

(a) The speed of the student at the bottom of the first hill can be found out using law of conservation of energy. The total energy at the top of the first hill is equal to the total energy at the bottom of the first hill.

The total energy at top of first hill is the sum of potential energy of the spring and potential energy of the student. Since the student was at rest there is no kinetic energy at the top of firect hill.

The total energy at the bottom of the hill is the kinetic energy of the student.

Equating both, we get

So the speed of the student when he reaches the bottom of the hil is 19.90m/s.

(b) Let h2 be the height the student reached up the second hill before momentarily stopping. Applying law of conservation of energy

So the student travelled a height of 20.20m up the hill.

If D is the distance he travelled, then

So the student travelled 40.40m up the hill before momentarily stopping.