The final speed of the crate is 6.3 m/s
Explanation:
Please note that there is a typo in the text of the question: the coefficient of sliding friction is 0.15, not 9.15.
First of all, we need to find the force of friction acting on the crate. This is given by:
[tex]F_f = \mu mg[/tex]
where:
[tex]\mu=0.15[/tex] is the coefficient of friction
m = 50 kg is the mass of the crate
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
Substituting,
[tex]F_f = (0.15)(50)(9.8)=73.5 N[/tex]
Now we can find the acceleration of the crate by using Newton's second law:
[tex]F-F_f = ma[/tex]
where
F = 90.0 N is the force applied forward on the crate
[tex]F_f = 73.5 N[/tex] is the force of friction, acting backward
a is the acceleration of the crate
Solving for a,
[tex]a=\frac{F-F_f}{m}=\frac{90.0-73.5}{50}=0.33 m/s^2[/tex] in the forward direction.
Finally, we can find the final speed of the crate by using suvat equations:
[tex]v^2-u^2 = 2as[/tex]
where:
v is the final speed
u = 0 is the initial speed
[tex]a=0.33 m/s^2[/tex] is the acceleration
s = 60 m is the distance covered
Solving for v,
[tex]v=\sqrt{u^2+2as}=\sqrt{0+2(0.33)(60)}=6.3 m/s[/tex]
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