To solve this problem, we can use the principle of conservation of mechanical energy. This principle states that the total mechanical energy (kinetic energy + potential energy) of an isolated system remains constant, as long as there are no non-conservative forces (like friction or air resistance) doing work on the system.
Given:
Mass of the ball (m) = 20 kg
Initial height (h1) = 100 m
Final height (h2) = 20 m
Height of the second hill (h3) = 30 m
Step-by-step Solution:
At the top of the first hill (height = 100 m): The ball starts stationary, so its initial kinetic energy is 0. Its potential energy is calculated using the formula: PE = mgh where:
m = 20 kg
g = 9.8 m/s² (acceleration due to gravity)
h = 100 m
So, initial potential energy (PE1) = 20 × 9.8 × 100 = 19600 J.
At the base of the hill (after rolling down): At the base of the hill, all the potential energy will have been converted into kinetic energy, as the ball is rolling without any friction (smooth surface).
The total mechanical energy is conserved, so the kinetic energy (KE) at the base will be equal to the initial potential energy. Therefore: KE = 19600 J.
The kinetic energy is given by the formula: KE = ½mv² where:
m = 20 kg
v = velocity of the ball at the base
Using the equation for kinetic energy: 19600 = ½ × 20 × v² Solving for v: v² = 19600 / 10 = 1960 v = √1960 ≈ 44.27 m/s.
At the top of the second hill (height = 30 m): When the ball climbs the second hill, some of its kinetic energy is converted back into potential energy. The potential energy at the top of this hill is: PE = mgh = 20 × 9.8 × 30 = 5880 J.
The remaining kinetic energy at this height is: KE_remaining = Initial kinetic energy - Potential energy at the top of the second hill KE_remaining = 19600 - 5880 = 13720 J.
Using the kinetic energy formula: KE_remaining = ½mv² 13720 = ½ × 20 × v² v² = 13720 / 10 = 1372 v ≈ √1372 ≈ 37.09 m/s.
At the base of the final hill (height = 20 m): Finally, the ball reaches a height of 20 m above the ground. The potential energy at this height is: PE = mgh = 20 × 9.8 × 20 = 3920 J.
The remaining kinetic energy is: KE_remaining = 13720 - 3920 = 9800 J.
Using the kinetic energy formula: KE_remaining = ½mv² 9800 = ½ × 20 × v² v² = 9800 / 10 = 980 v = √980 ≈ 31.3 m/s.
Thus, the velocity of the ball when it reaches the horizontal base at a height of 20 m is approximately 31.3 m/s.
However, if the problem states the velocity is 40 m/s, there may be a misunderstanding or an approximation in the setup of the problem. Based on the principles of conservation of mechanical energy and the information provided, the velocity at the horizontal base is around 31.3 m/s, not 40 m/s.
