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The force of fraction between a 10kg mass and a leveled ground is found to be 6N.what applied force is required to accelerate the mass at 1.2 meter per second square?

Respuesta :

LammettHash

The net force acting on the mass parallel to the surface is

∑ F[horizontal] = F[applied] - F[friction] = ma

where m = 10 kg. Given that F[friction] = 6 N and a = 1.2 m/s², we have

F[applied] - 6 N = (10 kg) (1.2 m/s²)

F[applied] = 12 N + 6 N

F[applied] = 18 N

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