Question

A 10kg monkey swings from one point which is 40m above the jungle floor to a point which is 15m above the floor. If the monkey was moving at 2m/s initially, what would be his speed at the 15m point?

Answer 1

The monkey's speed at the 15m point is approximately 22.2 m/s.

Here's how to solve the problem:

Step 1: Identify the relevant information:

Initial height (h1) = 40 m

Final height (h2) = 15 m

Initial speed (v1) = 2 m/s

Step 2: Choose the appropriate equation:

In this case, we can use the law of conservation of mechanical energy, which states that the total mechanical energy (sum of kinetic and potential energy) of a closed system remains constant.

Step 3: Apply the equation:

Initial total mechanical energy (E1) = Potential energy at h1 + Kinetic energy at v1

E1 = mgh1 + 1/2 mv1^2

Substituting the values: E1 = 10 kg * 9.8 m/s^2 * 40 m + 1/2 * 10 kg * (2 m/s)^2

E1 = 3920 J + 20 J

E1 = 3940 J

Final total mechanical energy (E2) = Potential energy at h2 + Kinetic energy at v2

E2 = mgh2 + 1/2 mv2^2

Substituting the values: E2 = 10 kg * 9.8 m/s^2 * 15 m + 1/2 * 10 kg * v2^2

E2 = 1470 J + 5v2^2

Step 4: Apply the conservation of energy principle:

E1 = E2

3940 J = 1470 J + 5v2^2

Step 5: Solve for v2:

2470 J = 5v2^2

v2^2 = 494 m^2/s^2

v2 = √494 m^2/s^2 ≈ 22.2 m/s

Therefore, the monkey's speed at the 15m point is approximately 22.2 m/s.