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
1
Expert's answer
2011-02-14T07:39:13-0500
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.
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