Answer to Question #1529 in Mechanics | Relativity for noor

Question

A 0.20 kg stone is held 2.0 m above the ground and then dropped .
1- find the stone's total mechanical energy at the point where it is dropped.
2- find the stone's total mechanical energy when it reached to the ground.
3- find the stone's speed at a height of 0.7m from the ground .
4- find the stone's speed when it reached to the ground .
5- find the height at which the stone's speed is 3.0 m/s

Answer 1

1. The total stone's mechanical energy is equal to its initial potential energy: mgh = 0.2 * 9.8 * 2.0 = 3.92 J.
2. Due to the law of energy concervation the total mechanical energy remains constant during the motion of the stone. E = 3.92 J.
3. According to the equations of the motion,
h =h₀ + v₀t + at²/2, v=v₀+ at the positive axis is directed upward, a=g is negative, v₀ = 0, h₀=2, h = 0.7. Let's find t.
2-0.7 = 9.8t2/2
t= 0.51 s.
V = 0.51*9.8 = 5.05 m/s.

4. h =h₀ - gt²/2, h=0:
2 = 9.8t2/2
t = 0.64 s
V = 0.64*9.8 = 6.26 m/s

5. 3 = 9.8t, t = 0.31s.
h =h₀ - gt²/2 = 2 - 9.8*0.312/2 = 1.54 m.