Answer:
The speed of the block after it has moved 3M if the surface in contact have a coefficient of kinetic friction of 0.15 is 1.7s m/sec
Explanation:
Given:
mass of the block = 6.0 kg
Force with which the block is pulled = 12 N
Kinetic friction \mu= 0.15
Distance travelled s = 3 m
To Find:
speed of the block after it has moved 3 metres =?
Solution :
W know that the friction formula is
[tex]f_k = \mu m g[/tex]
Substituting the values,
[tex]f_k = (0.15)(6)(10)[/tex]
[tex]f_k= 9 N[/tex]
Now Acceleration is Given by
[tex]a=\frac{F -f_k}{m}[/tex]
[tex]a=\frac{12 - 9}{60}[/tex]
[tex]a=\frac{3}{6}[/tex]
a= 0.5 [tex]m/s^2[/tex]
Initial velocity is u = 0
Also we know that,
[tex]v^2 - u^2=2as[/tex]
So the equation becomes
[tex]v^2 =2as[/tex]
[tex]v=\sqrt{2as}[/tex]
Substituting the values,
[tex]v=\sqrt{2as}[/tex]
[tex]v=\sqrt{2(0.5)(3)}[/tex]
[tex]v= \sqrt{3}[/tex]
v= 1.73 m/s
