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A 5.25-kg crate is suspended from the end of a short vertical rope of negligible mass. An upward force F(t) is applied to the end of the rope, and the height of the crate above its initial position is given by y(t)_________________.

Respuesta :

y(t) = 1/2 x F(t)/5.25 x t²

Explanation:

The upward force of F(t) is applied at the free end . Therefore the acceleration of  mass , suspended from rope can be calculated as

The displacement S = u t + 1/2 a t²

By substituting the values  S = 0 + 1/2 x F(t)/5.25 x t²

or  y(t) = 1/2 x F(t)/5.25 x t²

Here u is the initial velocity , because mass was at rest in the beginning , hence it is zero .

The acceleration a = force/mass = F(t)/5.25

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