Answer:
h =220 m
Explanation:
Given that
u = 7 m/s
Even mass will attach but this will not produce any effect on the maximum height of the ball.Because in energy conservation the effect of mass does not present.
So the final speed of the ball will be zero at the maximum height.
v² = u² - 2 g (25 + h)
0 = 7² - 2 x 10 (25 +h)
49 = 20 ( 25 +h)
49 = 500 +20 h
Here h comes out negative that is why we are taking the 70 m/s in place of 7 m/s.
0 = 70² - 2 x 10 (25 +h) ( take g =10 m/s²)
4900 = 20 ( 25 +h)
4900 = 500 +20 h
4900- 500 = 20 h
4400 = 20 h
440 = 2 h
h =220 m
If it attractive it has opposite pole and if it repulsive it has same pole
Answer:
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As we know that KE and PE is same at a given position
so we will have as a function of position given as
also the PE is given as function of position as
now it is given that
KE = PE
now we will have
so the position is 0.707 times of amplitude when KE and PE will be same
Part b)
KE of SHO at x = A/3
we can use the formula
now to find the fraction of kinetic energy
now since total energy is sum of KE and PE
so fraction of PE at the same position will be
