Question

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.What is the magnitude of the force F when 4.10s ?

Answer 1

Answer: 75.05N

Explanation:

According to newton's second law,

Force = mass×acceleration

Given mass = 5.0kg

Acceleration = change in velocity/time

Velocity = change in displacement/position/time i.e dy/dt

Given position of the particle as y(t)=(2.80 m/s)t +(0.61 m/s)t³

dy/dt = 2.80 + 3(0.61)t²

V = dy/dt = 2.80 + 1.83t²

Acceleration (a) = dv/dt = 2(1.83)t

dv/dt = 3.66t

To get the force when t =4.10second

dv/dt @ t = 4.10s will be 3.66(4.10)

acceleration = 3.66×4.10 = 15.01m/s²

Magnitude of the force F = ma

F = 5.0kg × 15.01m/s²

F = 75.05N