a steady force of friction acts on a 15.0-kg mass moving initially at 10.0 m/s on a horizontal surface. Show that if the mass is brought to rest over a distance of 12.5 m the friction force is 60 N

Respuesta :

As we know that block is moving at speed 10 m/s and comes to rest after travelling 12.5 m distance

so here we will have

[tex]v_f^2 - v_i^2 = 2a d[/tex]

[tex]0 - 10^2 = 2(a)(12.5)[/tex]

[tex]-100 = 25a[/tex]

[tex]a = -4 m/s^2[/tex]

now we know that

[tex]F = ma[/tex]

here we have

m = 15 kg

a = -4 m/s^2

so we will have

F = 15(4) = 60 N

so friction force on it will be 60 N