A steel ball of mass 50g is thrown vertically downwards with a velocity of 15m/s from a height of 20m . it buries itself in the sandy ground to a depth of 20cm . the magnitude of average force exerted on the ball by the sand is nearly: (g=10m/s2)
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Initial total energy = KE + PE
= 1/2 m v^2 + m g h
= 1/2 * 0.050 *15*15 + 0.050*10*20
= 5.625 + 10 = 15.625
Work done by the resisting force F in the sand = F * d
d = depth inside sand.
So F * 0.2 = 15.625
F = 78.125 N
= 1/2 m v^2 + m g h
= 1/2 * 0.050 *15*15 + 0.050*10*20
= 5.625 + 10 = 15.625
Work done by the resisting force F in the sand = F * d
d = depth inside sand.
So F * 0.2 = 15.625
F = 78.125 N
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