Respuesta :
Answer:
Final speed of the crate is 15 m/s
Explanation:
As we know that constant force F = 80 N is applied on the object for t = 12 s
Now we can use definition of force to find the speed after t = 12 s
[tex]F . t = m(v_f - v_i)[/tex]
so here we know that object is at rest initially so we have
[tex]80 (12) = 80( v_f - 0)[/tex]
[tex]v_f = 12 m/s[/tex]
Now for next 6 s the force decreases to ZERO linearly
so we can write the force equation as
[tex]F = 80 - \frac{40}{3} t[/tex]
now again by same equation we have
[tex]\int F .dt = m(v_f - v_i)[/tex]
[tex]\int (80 - (40/3)t) dt = 80(v_f - 12)[/tex]
[tex]80 t - \frac{40t^2}{6} = 80(v_f - 12)[/tex]
put t = 6 s
[tex]480 - 240 = 80(v_f - 12)[/tex]
[tex]v_f = 12 + 3[/tex]
[tex]v_f = 15 m/s[/tex]
The speed of the object is the motion of the object with respect to time.
The final speed of the crate after 18 seconds is 15 m/s.
How can you calculate the speed of crate?
Given that mass of the packing crate is 80 kg, and the applied force is 80 N for 12 seconds.
So force can be given as below.
[tex]f\times t =m v_1 -mv_2[/tex]
Where f is the force, t is time, m is the mass, v1 is the final velocity and v2 is the initial velocity.
[tex]80 \times 12 = 80\times v_1 -80\times 0[/tex]
[tex]v_1 =12 \;\rm m/s\\[/tex]
Hence the final velocity of the crate after 12 seconds is 12 m/s.
The force decreases to zero linearly for the next 6 seconds. Hence,
[tex]F = 80 - \dfrac {80}{6} t[/tex]
For the linear decrease in the value of force,
[tex]\int F dt = mv_1 - mv_2[/tex]
[tex]\int(80 - \dfrac{80}{6}t) dt = 80 v_1 -80\times 12[/tex]
[tex]80 t - \dfrac {80}{6\times 2}t^2 = 80v_1 - 960[/tex]
For the next 6 seconds, t = 6, then,
[tex]80\times 6 - \dfrac {80}{12}\times 6^2 = 80v_1 -960[/tex]
[tex]v_1 = 15 \;\rm m/s[/tex]
Hence, after 18 seconds, we can conclude that the final speed of the crate is 15 m/s.
To know more about the speed, follow the link given below.
https://brainly.com/question/862972.
