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mash [69]
3 years ago
10

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 en

d 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/s³)t³.
What is the magnitude of the force F when 4.10 s?
Is the magnitude's unit N but the system doesn't accept it?
Physics
1 answer:
Brut [27]3 years ago
7 0

Answer:

75 N

Explanation:

In this problem, the position of the crate at time t is given by

y(t)=2.80t+0.61t^3

The velocity of the crate vs time is given by the derivative of the position, so it is:

v(t)=y'(t)=\frac{d}{dt}(2.80t+0.61t^3)=2.80+1.83t^2

Similarly, the acceleration of the crate vs time is given by the derivative of the velocity, so it is:

a(t)=v'(t)=\frac{d}{dt}(2.80+1.83t^2)=3.66t [m/s^2]

According to Newton's second law of motion, the force acting on the crate is equal to the product between mass and acceleration, so:

F(t)=ma(t)

where

m = 5.00 kg is the mass of the crate

At t = 4.10 s, the acceleration of the crate is

a(4.10)=3.66\cdot 4.10 =15.0 m/s^2

And therefore, the force on the crate is:

F=ma=(5.00)(15.0)=75 N

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