A 5.00 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) = (2.80 m/s) t + (0.61 m/s^3) t^3.

(a) What is the magnitude of the force F when 4.10s?

(b) is the magnitude's unit N but the system doesn't accept it?

Question

Physics

Payten Arias

14 December, 13:27

A 5.00 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) = (2.80 m/s) t + (0.61 m/s^3) t^3.

(a) What is the magnitude of the force F when 4.10s?

(b) is the magnitude's unit N but the system doesn't accept it?

+4

Answers (1)

Know the Answer?

Cecelia Castillo · Answer 1

122.2

Explanation:

From the question, the equation of the force acting on the crate is,

F-mg = ma

F = ma+mg ... Equation 1

Where F = upward force, m = mass of the crate, a = acceleration of the crate, g = acceleration due to gravity of the crate.

Given: m = 5.0 kg, g = 9.8 m/s²

And a,

a = d²y/dt²

Where,

y (t) = 2.8t+0.61t³

d²y/dt² = 3.66t ... equation 2

when t = 4.0 s.

substitute into equation 2

d²y/dt² = a = 3.66 (4)

a = 14.64 m/s²

substitute into equation 1,

F = 5 (14.64) + 5 (9.8)

F = 73.2+49

F = 122.2 N.