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Wants to slide his 430g water bottle precisely 12m to land inside a Bull's Eye! If the coefficient of kinetic friction between the bottle and the surface is 0.62, calculate the initial speed he must release the bottle?

Respuesta :

Answer:

Ff = u M g    frictional force where u = .62

1/2 M v^2    kinetic energy of water bottle at release

Ff * d = 1/2 M v^2 = u M g d   work to stop equals initial kinetic energy

v^2 = 2 u g d = 2 * .62 * 9.8 * 12 = 146 m^2 / s^2

v = 12.1 m/s

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