Question

:1. A 20-kg sled has gotten away. The sled starts from rest and goes down a straight down the hill. How fast will be the sled be at the bottom of the hill if it loses 100-m in altitude and friction is negligible?
2. Suppose that the sled in the previous example is observed to be 30 m/s at the bottom of the hill. How much energy is converted to thermal energy by friction on the way down?

Answer 1

Question #76720, Physics / Mechanics | Relativity

1. A 20-kg sled has gotten away. The sled starts from rest and goes down a straight down the hill. How fast will be the sled be at the bottom of the hill if it loses 100-m in altitude and friction is negligible?

Solution

From the conservation of energy:

$$\frac{m v^{2}}{2} = m g h$$ $$v = \sqrt{2 g h} = \sqrt{2 (9.8) (100)} = 44 \frac{m}{s}.$$

Answer: $44 \frac{m}{s}$.

2. Suppose that the sled in the previous example is observed to be $30\,\mathrm{m/s}$ at the bottom of the hill. How much energy is converted to thermal energy by friction on the way down?

Solution

From the conservation of energy:

$$m g h = \frac{m v^{2}}{2} + Q$$ $$Q = m \left(g h - \frac{v^{2}}{2}\right) = 20 \left((9.8) (100) - \frac{30^{2}}{2}\right) = 11000\,J = 11\,kJ.$$

Answer: $11\,kJ$

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