Question

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)_________________.

Answer 1

In Physics, the tension in the rope holding a 5.25-kg stationary crate equals the weight of the crate, calculated using Newton's second law with no acceleration involved.

Step-by-step explanation:

The subject of this question is Physics, specifically relating to Newton's laws of motion and simple harmonic motion. For a 5.25-kg crate that is hanging stationary from a rope, the tension in the rope must balance the weight of the crate. This can be shown using Newton's second law, where the net force (Fnet) acting on the crate is zero because the crate is not accelerating. Therefore, the tension T in the rope is equal to the weight of the crate (w), which can be calculated using the equation T = w = mg, where m is the mass of the crate and g is the acceleration due to gravity (approximately 9.80 m/s²). For a 5.25-kg crate, the tension would be T = (5.25 kg) × (9.80 m/s²) = 51.45 N. This concept also applies to scenarios involving simple harmonic motion (SHM), where a mass attached to a spring oscillates around an equilibrium position.

Answer 2

y(t) = 1/2 x F(t)/5.25 x t²

Step-by-step 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